50 AMC Lectures Chapter 26 Counting BASIC KNOWLEDGE … · 50 AMC Lectures Chapter 26 Counting BASIC ... objects in which the order is important. For example, if we have three numbers

50 AMC Lectures Chapter 26 Counting

BASIC KNOWLEDGE 1. Two Important Terms (1.1). Permutations A permutation is an arrangement or a listing of objects in which the order is important. For example, if we have three numbers 1, 5, 9, there are 6 total permutations: {1,5,9}, {1,9,5}, {5,1,9}, {5,9,1}, {9,1,5}, {9,5,1}. (1). We have n different elements, and we would like to arrange r of these elements with no repetition, where 1 ≤ r ≤ n. The number of such permutations is

)!(

!),(rn

nrnP−

= (1.1)

(2). We have n different elements, and we would like to arrange all n of these elements with no repetition. We let r = n in (1) to get P(n, n) = n! (1.2) These n distinct objects can be permutated in n! permutations. The symbol ! (factorial) is defined as follows:

0! = 1 (1.3) and for integers n ≥ 1,

n!= n · (n – 1) ···· 1 (1.4) 1! = 1 2! = 2 · 1= 2 3! = 3 · 2 · 1= 6 4! = 4 · 3 · 2 · 1 = 24 5! = 5 · 4 · 3 · 1 · 1 = 120 6! = 6 · 5 · 4 · 3 · 2 · 1 = 720

1


Proof of (1.2): The first object can be chosen in n ways, the second object in n −1 ways, the third in n − 2, etc. By the Fundamental Counting Principle, we have n(n − 1)(n − 2) · · · 2 · 1 = n! ways. Example 1: How many 5-digit positive integers can be formed by the digits of 0, 1, 2, 3, and 4? Solution: Since 12340 is different from 13240, the order in which the digits are arranged is important, so we can use permutations to solve this problem. Method 1: The number of permutations of 5 digits is 5! = 120. However, we have to also consider the permutations of 5 digits where 0 is the leftmost digit. Because 0 cannot be the leftmost digit, these permutations must be subtracted from the total permutations of 5 digits. The number of permutations of 5 digits, with 0 as in the leftmost digit, is 4! = 24. No positive 5-digit integers can be formed by these permutations, so the answer is then 120 – 24 = 96. Method 2: The number of permutations with 0 as the units digit is 4! = 24. The number of permutations with 0 as the tens digit is 4! = 24. The number of permutations with 0 as the hundred digit is 4! = 24. The number of permutations with 0 as the thousand digit is 4! = 24. The answer is then 24 × 4 = 96. (1.2). Combinations Definition: A combination is an arrangement or a listing of things in which order is not important.

Let n, r be non-negative integers such that 0 ≤ r ≤ n. The symbol (read “n choose m”)

is defined and denoted by

⎟⎟⎠

⎞⎜⎜⎝

⎛rn

2


)!(!!

),(),(

rnrn

rrPrnP

rn

−==⎟⎟

⎠

⎞⎜⎜⎝

⎛ (1.5)

Remember: , , and 1 nn

=⎟⎟⎠

⎞⎜⎜⎝

⎛10

=⎟⎟⎠

⎞⎜⎜⎝

⎛n1=⎟⎟

⎠

⎞⎜⎜⎝

⎛nn

Since n – (n – r) = r, we have (1.6) ⎟⎟⎠

⎞⎜⎜⎝

⎛−

=⎟⎟⎠

⎞⎜⎜⎝

⎛rn

nrn

Unlike permutations, combinations are used when the order of the terms does not matter. If we have n different elements, and it doesn’t matter which order we arrange the

elements, the number of combinations to arrange m elements where 1 ≤ m ≤ n, is ⎜⎜ ⎟⎟⎠

⎞mn

⎝

⎛

⎟⎟⎠

⎞

Example 2: (a). In how many parts at most do n lines cut a plane? (b). In how many parts at most do n planes cut a space? Solution:

(a). ⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛210nnn

(b). ⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛3210nnnn

2. Two Important Rules (2.1). The product rule (Fundamental Counting Principle) (Step Work) When a task consists of k separate steps, if the first step can be done in n1 ways, the second step can be done in n2 ways, and so on through the kth step, which can be done in nk ways, then the total number of possible results for completing the task is given by the product: (2.1) knnnnN ××××= ...321

Example 3: (1998 North Carolina Math Contest) There are 8 girls and 6 boys in the Math Club at Central High School. The Club needs to form a delegation to send to a conference, and the delegation must contain exactly two girls and two boys. The number of possible delegations that can be formed from the membership of the Club is a) 48 b) 420 c) 576 d) 1680 e) 2304

3


Solution: (B). In order to form the delegation, we need to go through the following two steps: Step 1: Selecting two girls:

The number of combinations to select two girls out of eight is . ⎟⎟⎠

⎞⎜⎜⎝

⎛28

Step 2: Selecting two boys:

The number of combinations to select two boys out of six is . ⎟⎟⎠

⎞⎜⎜⎝

⎛26

By the product rule, we have the total number of distinct delegations:

4

⎟⎟⎠

⎞⎟⎟⎠

⎞⎜⎜⎝

⎛28

× = 28 × 15 = 420. ⎜⎜⎝

⎛26

(2.2). The sum rule (case work) If an event E1 can happen in n1 ways, event E2 can happen in n2 ways, event Ek can happen in nk ways, and if any event E1, E2,.. or Ek happens, the job is done, then the total ways to do the job is N = n1 + n2 + ···+ nk (2.2) Example 4: Hope High School has three elective courses for social studies and four electives for science. How many ways are there for Alex to select three electives from them this semester? (A) 30 (B) 35 (C) 42 (D) 48 Solution: (B). Case 1: Alex can select 1 social studies course and 2 science courses. By the product rule, there are

1824

13

=⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟⎟⎠

⎞⎜⎜⎝

⎛ ways for Alex to select 1 social studies course and 2 science courses.

Case 2: Alex can select 2 social studies courses and 1 science course. By the product rule, there are


1214

23

=⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟⎟⎠

⎞⎜⎜⎝

⎛ways for Alex to select 2 social studies courses and 1 science course.

Case 3: Alex can select 3 social studies courses in ways. 333

=⎟⎟⎠

⎞⎜⎜⎝

⎛

Case 4: Alex can select 3 science courses in ways. 434

=⎟⎟⎠

⎞⎜⎜⎝

⎛

By the sum rule, there are 18 + 12 + 1 + 4 = 35 total ways. Example 5: How many two-digit numbers are there such that the units digit is greater than the tens digit? Solution: We want to count the number of two-digit number that have a larger units digit than tens digit, giving us the following 4 cases: Case I: When the units digit is 9, the tens digit can be 1, 2, 3, 4, 5, 6, 7 or 8, so we have 8 such two-digit numbers. Case II: When the units digit is 8, the tens digit can be 1, 2, 3, 4, 5, 6 or 7, so we have 7 such two-digit numbers. Case III: When the units digit is 7, the tens digit can be 1, 2, 3, 4, 5, or 6, so we have 6 such two-digit numbers. Case IV: When the units digit is 6, the tens digit can be 1, 2, 3, 4, or 5, so we have 5 such two-digit numbers. Case V: When the units digit is 5, the tens digit can be 1, 2, 3, or 4, so we have 4 such two-digit numbers. Case VI: When the units digit is 4, the tens digit can be 1, 2, or 3, so we have 3 such two-digit numbers. Case VII: When the units digit is 3, the tens digit can be 1 or 2, so we have 2 such two-digit numbers.

5


Case VIII: When the units digit is 2, the tens digit can be 1 so we have 1 such two-digit number.

By The Sum Rule, we have 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36942

8)81(=×=

×+ such

two-digit numbers. Example 6: (1989 AMC) Mr. and Mrs. Zeta want to name baby Zeta so that its monogram (first, middle, and last initials) will be in alphabetical order with no letters repeated. How many such monograms are possible? (A) 276 (B) 300 (C) 552 (D) 600 (E) 15600 Solution: (B). The last initial is fixed at Z. If the first initial is A, the second initial must be one of B, C, D, . . ., Y, so there are 24 choices for the second. If the first initial is B, there are 23 choices for the second initial: of C, D, E, . . ., Y. Continuing in this way we see that the number of monograms is 24 + 23 + 22 + · · · + 1.

Using the formula 2

)1(21 +=+++

nnnL , we get the answer .3002

2524=

⋅

Example 7: (2003 AMC 10B) How many distinct four-digit numbers are divisible by 3 and have 23 as their last two digits? Solution: A number is divisible by 3 if the sum of its digits are divisible by 3; therefore, the first two digits should have a sum of 1, 4, 7, 10, 13, or 16. If the sum of the first two digits is 1, there is only 1 such number: 1023.

If the sum of the first two digits is 4, there are 4 such numbers because

4 = 4 + 0 = 1 + 3 = 3 + 1 = 2 + 2.


7 = 7 + 0 = 6 + 1 = 1 + 6 = 5 + 2 = 2 + 5 = 4 + 3 = 3 + 4.


6


9 = 9 + 0 = 8 + 1 = 1 + 8 = 7 + 2 = 2 + 7 = 6 + 3 = 3 + 6 = 5 + 4= 4 + 5.


13 = 9 + 4 = 4 + 9 = 8 + 5 =5 + 8 = 7 + 6 = 6 + 7.


16 = 9 + 7 = 7 + 9 = 8 + 8.

By the sum rule, we get the answer: 1 + 4 + 7 + 9 + 6 + 3 = 30.

Example 8: (1989 AIME) Ten points are marked on a circle. How many distinct convex polygons of three or more sides can be drawn using some (or all) of the ten points as vertices? (Polygons are distinct unless they have exactly the same vertices.) Solution: 968. We would like to form convex polygons. We have the following cases, and in each case, we finish the job of forming convex polygons.

7

⎟⎟⎠

⎞Triangles: The number of triangles formed by selecting three points out of 10 points is

⎜⎜⎝

⎛310

Convex quadrilaterals: . ⎟⎟⎠

⎞⎜⎜⎝

⎛410

Convex pentagons: . ⎟⎟⎠

⎞⎜⎜⎝

⎛510

Convex hexagons: . ⎟⎟⎠

⎞⎜⎜⎝

⎛610

Convex heptagons: . ⎟⎟⎠

⎞⎜⎜⎝

⎛710

Convex octagons: . ⎟⎟⎠

⎞⎜⎜⎝

⎛810

Convex nonagons: . ⎟⎟⎠

⎞⎜⎜⎝

⎛910


Convex decagons: . ⎟⎟⎠

⎞⎜⎜⎝

⎛1010

By the sum rule, the number of convex polygons is

8

⎟⎟⎠

⎞⎟⎟⎠

⎞⎜⎜⎝

⎛

⎝N = + ⎜⎜ + + + + + + = 210 – ( + ⎜⎜

⎛+

) = 1024 – 56 = 968.

⎜⎜⎝

⎛310

⎟⎟⎠

⎞210

⎝

⎛410

⎟⎟⎠

⎞⎜⎜⎝

⎛510

⎟⎟⎠

⎞⎜⎜⎝

⎛610

⎟⎟⎠

⎞⎜⎜⎝

⎛710

⎟⎟⎠

⎞⎜⎜⎝

⎛810

⎟⎟⎠

⎞⎜⎜⎝

⎛910

⎟⎟⎠

⎞⎜⎜⎝

⎛1010

⎟⎟⎠

⎞010

⎟⎟⎠

⎞110

⎜⎜⎝

⎛

3. Three Important Theorems THEOREM 1: (Grouping) (a). Let the number of different objects be n. Divide n into r groups A1, A2, ..., Ar such that there are n1 objects in group A1, n2 objects in group A2, ..., nr objects in the group Ar, where n1 + n2 + · · · + nr = n. The number of ways to do so is

!

⎟⎟⎠

⎞

⎟⎟⎠

⎞

⎟⎟⎠

⎞

!!!

21 rnnnnN⋅⋅⋅

= (3.1)

Proof:

There are ways to take out n1 elements from n elements to form group A1. ⎜⎜⎝

⎛

1nn

There are ways to take out n2 elements from n – n1 elements to form group A2. ⎜⎜⎝

⎛ −

2

1

nnn

Continue the process until there are nr elements left to form group Ar. The total number of ways, based on the Fundamental Counting Principle, is

⎟⎟⎠

⎞⎜⎜⎝

⎛ −

2

1

nnn

⎜⎜⎝

⎛

1nn

.... = ⎟⎟⎠

⎞⎜⎜⎝

⎛

r

r

nn

!!!!

21 rnnnn⋅⋅⋅

(b). Let there be r types of objects: n1 of type 1, n2 of type 2; etc. The number of ways in which these n1 + n2 + · · · + nr = n objects can be rearranged is

!!!!

21 rnnnn⋅⋅⋅

(3.2)

The proof of this is the same as the proof for (3)


Example 9: A gardener plants eight trees out of three maple trees, two oak trees, and four birch trees in a row. How many ways are there? Solution: Case I: With two maple trees, two oak trees, and four birch trees, by (3.1), we have

420!4!2!2

!8= ways.

Case II: With three maple trees, one oak tree, and four birch trees, there are

280!4!1!3

!8= ways.

Case III: With three maple trees, two oak trees, and three birch trees, there are

560!3!2!3

!8= ways.

By the sum rule, we know that the total number of ways to plant the trees is 420 + 280 + 560 = 1260. Example 10: Five numbers 1, 2, 3, 4, and 5 are arranged in a row like . How many arrangements are there such that a1 ≠ 1, a2 ≠ 2, a3 ≠ 3, a4 ≠ 4, a5 ≠ 5 ?

54321 aaaaa

Solution: We let 2 be the leftmost number. As shown in the figure to the right, there are 11 such arrangements: Since the leftmost digit can also be 3, 4, or 5, so the answer by the product rule will be 11 × 4 = 44. THEOREM 2: (Combinations with Repetitions) (a). n identical balls are put into r labeled boxes and the number of balls in each box is not limited. The number of ways is

9


10

⎟⎟⎠

⎞⎜⎜⎝

⎛ −+n

rn

1 or (3.3) ⎟⎟

⎠

⎞⎜⎜⎝

⎛−−+1

1r

rn

Proof: Put r labeled boxes next to each other as shown in the figure below. Put n balls into these boxes (Figure 1). Next, we line these boxes up next to each other (Figure 2). Now we take apart the top and bottom sides of the each box and the two sides of the two boxes at the end (Figure 3), resulting figure 4. The problem now becomes finding the number of ways to permute n identical balls with r

– 1 identical partitions: )!1(!)!1(

−−+

rnrn or or . ⎟⎟

⎠

⎞⎜⎜⎝

⎛ −+n

rn

1⎟⎟⎠

⎞⎜⎜⎝

⎛−−+1

1r

rn

Figure 1 Figure 2

Figure 3 Figure 4 (b) The number of terms in the expansion of , after the like terms combined, is

nrxxxx )....( 321 ++++

⎟⎟⎠

⎞

⎟⎟⎠

⎞

⎜⎜⎝

⎛ −+n

rn

1 or (3.4) ⎟⎟

⎠

⎞⎜⎜⎝

⎛−−+1

1r

rn

(c). Let n be a positive integer. The number of positive integer solutions to x1 + x2 + ⋅⋅⋅ + xr = n is

⎜⎜⎝

⎛−−

11

rn

. (3.5)


Proof:

11

⎟⎟⎠

⎞

⎟⎟⎠

⎞

Write n as n =1+1+ ⋅⋅⋅ +1+1, where there are n 1’s and n − 1 plus signs. In order to decompose n into r summands, we choose r − 1 plus signs from the n − 1, giving us

ways to do so. ⎜⎜⎝

⎛−−

11

rn

(d). Let n be a positive integer. The number of non-negative integer solutions to y1 + y2 + ⋅⋅⋅ + yr = n is

⎜⎜⎝

⎛ −+n

rn

1 or (3.6) ⎟⎟

⎠

⎞⎜⎜⎝

⎛−−+1

1r

rn

Proof: Set xr − 1= yr. Then xr ≥ 1. The equation x1 + x2 + ⋅⋅⋅ + xr = n

is equivalent to x1 + x2 + ⋅⋅⋅ + xr = n + r, which has solutions. ⎟⎟⎠

⎞⎜⎜⎝

⎛−−+1

1r

rn

Example 11: A baking company produces four different cookies: Chocolate Chip Cookies, Peanut Butter Cookies, Oatmeal Cookies, and Blueberry Cookies. (a) If a package contains 8 cookies, how many different packages are possible? (b) If a package contains 8 cookies with at least one cookie of each kind, how many different packages are possible? Solution: (a). From (3.6), we write y1 + y2 + ⋅⋅⋅ + yr = n ⇒ y1 + y2 + y3 + y4 = 8

There are different packages possible. 165311

3 148

1 1

=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛ −+=⎟⎟

⎠

⎞⎜⎜⎝

⎛−−+

=r

rnN

(b). By (3.5), we write y1 + y2 + ⋅⋅⋅ + yr = n ⇒ y1 + y2 + y3 + y4 = 8

There are different packages possible. 3537

1418

11

=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛−−

=⎟⎟⎠

⎞⎜⎜⎝

⎛−−

=rn

N

Example 12: (1998 AIME) Find the number of ordered quadruples (x1 , x2 , x3 , x4) of positive odd integers that satisfy x1 + x2 + x3 + x4 = 98.


Solution: Since x1 , x2 , x3 , x4 are odd positive integers, we let x1 = 2y1 − 1 x2 = 2y2 − 1 x3 = 2y3 − 1 x4 = 2y4 − 1 y1, y2, y3, and y4 are positive integers. The given equation becomes: 2(y1 + y2 + y3 + y4) = 98 + 4 ⇒ y1 + y2 + y3 + y4 = 51 (1) By (3.5), the number of positive integer solutions is

19600350

14151

=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛−−

.

Example 13: (a) How many ways are there to take 4 letters from a, b, b, c, c, c, d, d, d, d, d ?

(b) How many different 4-letter codes can be formed by using the letters from a, b, b, c, c, c, d, d, d, d, d ?

Solution: (a) 20, (b) 152.

Method 1:

Case 1: 4 letters are the same: . 111

=⎟⎟⎠

⎞⎜⎜⎝

⎛


12

=⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟⎟⎠

⎞⎜⎜⎝

⎛

12

⎟⎟⎠

⎞

⎟⎟⎠

⎞

⎜⎜⎝

⎛12

represents the number of ways to select the 3 same letters from either the letters c or

d.

⎜⎜⎝

⎛13

denotes the number of ways to select the final letter from a, b, or the letter (c or d)

that was not selected before.


Case 3: 2 letters are the same and 2 other letters are the same: 32

12

13

=⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟⎟⎠

⎞⎜⎜⎝

⎛

Case 4: 2 letters are the same, 2 other letters are different: 913

13

=⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟⎟⎠

⎞⎜⎜⎝

⎛

Case 5: 4 letters are different: 144

=⎟⎟⎠

⎞⎜⎜⎝

⎛

Total 1 + 6 + 3 + 9 + 1 = 20.

Method 2 (by listing): Case 1: 4 letters are the same: dddd. We have 1 way for case 1.

Case 2: 3 letters are the same: ccca 1 way. cccb 1 way cccd 1 way ddda 1 way dddb 1 way dddc 1 way We have 6 ways for case 2. Case 3: 2 letters are the same and 2 other letters are the same: bbcc 1 way bbdd 1 way ccdd 1 way We have 3 ways for case 3. Case 4: 2 letters are the same, 2 other letters are different: bbac 1 way bbad 1 way bbcd 1 way ccab 1 way ccad 1 way ccbd 1 way

13


ddab 1 way ddac 1 way ddbd 1 way We have 9 ways for case 4. Case 5: 4 letters are different: abcd. We have 1 way for case 5. Total 1 + 6 + 3 + 9 + 1 = 20.

(b) 152.

Method 1:


=⎟⎟⎠

⎞⎜⎜⎝

⎛

Case 2: 3 letters are the same: 24!3!4

13

12

=×⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟⎟⎠

⎞⎜⎜⎝

⎛.

⎟⎟⎠

⎞

⎟⎟⎠

⎞

⎜⎜⎝

⎛12

represents the number of ways we select the 3 same letters from either the letters c

or d.

⎜⎜⎝

⎛13

denotes the number of ways to select the final letter from a, b, or the letter (c or d)

that was not selected before.

!3!4 is the number of ways to arrange the 3 same letters with the fourth letter.

Case 3: 2 letters are the same and 2 other letters are the same: 18!2!2

!423

=×

×⎟⎟⎠

⎞⎜⎜⎝

⎛

Case 4: 2 letters are the same, 2 other letters are different: 108!2!4

23

13

=×⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟⎟⎠

⎞⎜⎜⎝

⎛

Case 5: 4 letters are different: 144

=⎟⎟⎠

⎞⎜⎜⎝

⎛

Total 1 + 24 + 18 + 108 + 1 = 152.

14


Method 2 (by listing): Case 1: 4 letters are the same: dddd. We have 1 way for case 1.

Case 2: 3 letters are the same:

ccca 4!3!4= ways

cccb 4 ways cccd 4 ways ddda 4 ways dddb 4 ways dddc 4 ways We have 24 ways for case 2. Case 3: 2 letters are the same and 2 other letters are the same:

bbcc 16!2!2

!4=

×

bbdd 6 ways ccdd 6 ways We have 18 ways for case 3. Case 4: 2 letters are the same, 2 other letters are different:

bbac 12!2!4=

bbad 12 ways bbcd 12 ways cc ab 12 ways ccad 12 ways ccbd 12 ways ddab 12 ways ddac 12 ways ddbd 12 ways We have 108 ways for case 4. Case 5: 4 letters are different: abcd

15


We have 1 way for case 5. Total 1 + 24 + 18 + 108 + 1 = 152. THEOREM 3: (Circular Permutations) The number of circular permutations (arrangements in a circle) of n distinct objects is N = (n – 1)! (3.7) We can think of this as n people being seated at a round table. Since a rotation of the table does not change an arrangement, we can put person A in one fixed place and then consider the number of ways to seat all the others. Person B can be treated as the first person to seat and M the last person to seat. The number of ways to arrange persons A to M is the same as the number of ways to arrange persons B to M in a row. So the number of ways to seat n people around a round table, or arranging n distinct objects around a circle, is N = (n – 1)!.

Example 14: In how many ways is it possible to seat seven people at a round table if Alex and Bob must not sit in adjacent seats? Solution: 480. By (3.7), the number of ways to sit 7 people at a round table is N = (n – 1)! = N = (7 – 1)! = 6! = 720. We can figure out the number of ways that Alex and Bob sit together by seeing them as one single entity, giving us (6 – 1)! = 5! ways where Alex and Bob are sitting next to each other. This is multiplied by 2 since we can alter Alex’s and Bob’s positions. The number of ways to seat seven people at a round table if Alex and Bob must not sit in adjacent seats is then 720 – 2 × 5! = 720 – 2 × 120 = 480. Example 15: In how many ways can four married couples be seated at a round table if no two men, as well as no husband and wife are to be in adjacent seats?

16


Solution: 12. There are (4 – 1)! = 3! to seat four women. After the ladies are seated, person M4 (whose wife is not shown in the figure below) has two ways to sit. After he is seated in any one of the two possible seats, the other men have only one way to sit in the remaining seats. The solution is 3! × 2 = 12.

Example 16: Twelve student body members are seated at a round table electing president, vice president, and treasurer. How many possible ways are there such that at least two of the three elected had been sitting next to each other? Solution: Case I: Three had been sitting next to each other. We have 12 ways to select first person, say, A. Once A is selected, we have one way to select B and C. By the product rule, the number of ways is 12 × 1 × 1 = 12.

Case II: Two of them had been sitting next to each other. We have 12 ways to select two that had been sitting next to each other. Let us say, A and B. For the third person, we are not allowed to select neither C nor D. So we have 12 – 4 ways to select the third person. By the product rule, number of ways is 12 × (12 – 4) = 96.

Finally by the sum rule, we have the answer: 12 + 96 = 108.

17


18

. Typical Problems in Counting4

.1 Counting problems related to partitioning 4

udents. In

how ma their chairs? (A) 12 (B) 36 (C) 60 (D) 84 (E) 630

olution:

t. Once the stu

professors to choose (two spaces at the ends do not count because each professor

tween two students) and there are 3! ways for them to rearrange emselves.

Example 17: (1994 AMC) Nine chairs in a row are to be occupied by six students and Professors Alpha, Beta and Gamma. These three professors arrive before the six studentsand decide to choose their chairs so that each professor will be between two st

ny ways can Professors Alpha, Beta and Gamma choose

S We sit the students firs dents are seated, there are five spaces for three

⎟⎠

⎜⎝

⎛3

needs to sit be

⎟⎞

⎜5

th

By the product rule, the answer is × 3! = 60.

s not to sit next to each other.

here are n arrangements and the last two digits of n is

）040 （B）520 （C）440 （D）720

olution: (A).

an first seat the etween the students.

⎟⎠

⎜⎝3

Example 18: Ten chairs in a row are to be occupied by eight students and two teacherAlpha and Beta for a class picture. Two teachers decide

⎟⎞

⎜⎛5

T A S Because the two teachers decide to not sit next to each other, we cstudents, and then seat the two teachers in bThere are 8! ways to eat the students first.


19

After that, we see that there are 9 spaces for two teachers to sit, giving us ⎜⎜⎝

⎛ ways to

seat them, and we must also note that there are 2! ways for them to rearrange themselves.

⎟⎟⎠

⎞29

By the product rule, the total number of ways to seat the teachers and students is 8

⎟

! ×

⎟⎞⎛9

× 2! = 40320 × 72 = 2903040. ⎠⎝2

⎜⎜

4.2. Counting problems related to coloring Example 19: As show color. There are 5 colo

n in the figure below, each of five regions ABCDE is to be assigned rs to choose from, and no adjacent regions can be the same color.

d more than once?

B (we

xam A show gure elow, re 4 colors to choose from, and no adjacent regions canfferent ways are there if each color is allowed to use mo

C．60 D．

ase II: A and C have different colors.

aHow many different ways are there if each color is allowed to be use Solution: 540 ways.

e have 5 ways to color region A first. Then we color regionWhave 4 ways). Next we color region C (we have 3 ways). For region D,

e have 3 ways (not the same color with A or C). Finally we have 3 wways to color E (not the same color with A or D). By the product rule, we have 5 × 4 × 3 × 3 × 3 = 540 different ways to color. E ple 20: s n in the fi b each of six regions ABCD is to be assigned a

be the same color. re than once?

48

color. There aow many diH

A．96 B．84

olution: B. S Case I: A and C have the same color.

ix regions is 4 × 3 × 1 × 3 = 36. The number of ways to color the s C


20

s of five sides of convex pentagon ABCDE are distinct. Each de is to be assigned a color. There are 3 colors (red, yellow and blue) to choose from,

sharing the same vertex must have different colors. How many different olorings are possible?

Solution: 30. We label the sides as follows:

The number of ways to color the six regions is 4 × 3 × 2 × 2 = 48. The answer is 36 + 48 = 84 ways. Example 21: The lengthsiand two sides c

If a is colored red, b is colored yellow, then when c is colored red, as shown in the figure below:

we have two arrangements:

If a is colored red, b is colored yellow, then when c is colored blue, as shown in the figure below:


we have three arrangements:

By the sum rule, we have 2 + 3 = 5 ways. We have 5 total ways if (a,b) is colored (red, yellow). We see that (a, b) can also colored (red, blue), (yellow, blue), (yellow, red), (blue, red), and (blue, yellow). Therefore the total number of colorings is 5 × 6 = 30. Example 22: Each of four faces of a regular tetrahedron is colored one of 10 colors. How many distinct ways are there to color the tetrahedron? (Two colorings are considered distinct if they cannot be rotated to look like each other). (A) 925 (B) 980 (C) 1024 (D) 1090 (E)1450 Solution: Case I: Four colors are used: We have two configurations as shown in the figure to the right:

21


The number of ways to color is

Case II: Three colors are used:

We have three configurations as shown in the figure to the right:


Case III: Two colors are used: We have three configurations as shown in the figure to the right: The number of ways to color is

Case IV: One color is used: We have only one configuration as shown in the figure to the right:


Total ways: 420 + 360 + 135 + 10 = 925

Example 23: (2003 North Carolina Math Contest) The tips of a five-pointed star are to be painted red, white and blue. How many ways can this be done if no adjacent points can be the same color? Solution: 30 ways. Method 1 (official solution): Let the 5 points be ABCDE. Note that to paint the 5 points in the prescribed fashion, 2 colors must be used

22


twice and one color just once. Also, note that if point A is a fixed color, then there are only two ways to paint the star. Consider one such set of colors, where red is the single color used. Then if A is painted red there are only two ways to paint the rest of the star. A similar situation occurs if the points B, C, D, or E are painted with the red. For this set of colors there are 10 ways to paint the star. Since there are three different sets of colors, there are a total of 30 ways to paint the star. Method 2 (our solution): If A and C have the same color, then the number ways to color will be:

A B C D E 3 × 2 × 1 × 1 × 2 = 12

If A and D have the same color, then the number of ways to color will be:

A B C D E 3 × 2 × 1 × 1 × 2 = 12

If A and C have the different color, then the number of ways to color will be:

A B C D E 3 × 2 × 1 × 1 × 1 = 6

Total ways is 12 + 12 + 6 = 30.

23


Method 3: There is a formula for this problem which can be derived by the recursion method. An = (m – 1)[(m – 1)(n-1) + (–1)n], where n is the number of points and m is the number of colors. In the case for the NC Math contest problem, n = 5 and m = 3. An = (m – 1)[(m – 1)(n-1) + (–1)n] = (3 – 1)[(3 – 1)(5 – 1) + (–1)5] = 30. Example 24. There are five regions to be colored with four different colors. If no same color can be used for adjacent regions, how many ways are there to color? Solution: If regions 1 and 4 are colored by the same color, we have the following ways to do so: Regions 1 2 3 4 5

Ways to color 4 × 3 × 2 × 1 × 2 = 48 If regions 1 and 4 are colored by different colors, we have the following ways to do so: Regions 1 2 3 4 5

Ways to color 4 × 3 × 2 × 1 × 1 = 24

The total number of ways is 48 + 24 = 72. Example 25. Each point of A, B, C, D, E, and F is to be assigned a color. There are 4 colors to choose from, and the ends of each line segment must have different colors. How many different colorings are possible? （A） 288 （B）264 （C） 240 （D）168 Solution: B. Case I：4 colors are used for B, D, E, and F. We have ways. 4

4 1 1 2A × × = 4 Case II: 3 colors are used for B, D, E, and F. We have ways. 3

4 2 2A × × + 34 2 1 2 192A × × × =

24


Case III: 2 colors are used for B, D, E, and F. We have ways. 2

4 2 2 4A × × = 8By the product rule, we have total 24 + 192 + 48 = 264 ways. 4.3. Counting problems related to geometry Example 26. (1993 AMC) How many triangles with positive area are there whose vertices are points in the xy-plane whose coordinates are integers (x, y) satisfying 1 ≤ x ≤ 4 and 1 ≤ y ≤ 4? Solution: 516.

There are = 560 sets of 3 points. We must exclude from our count those sets of

three points that are collinear.

⎟⎟⎠

⎞

⎟⎟⎠

⎞

⎜⎜⎝

⎛316

There are 4 vertical and 4 horizontal lines with four points each. These 8 lines contain

8 ⎜⎜ = 32 sets of 3 collinear points. Similarly, there are 2 ⎜⎜ + 4 ⎜⎜ = 12 sets of 3

collinear points that determine lines of slope ±1. Because there are no other sets of 3 collinear points, the number of triangles is the total sets minus the degenerate triangles equals 560 – 32 – 12 = 516.

⎝

⎛34

⎟⎟⎠

⎞

⎝

⎛34

⎟⎟⎠

⎞

⎝

⎛33

25


Example 27. (1976 AMC) Lines L1, L2. . ., L100 are distinct. All lines L4n, n a positive integer, are parallel to each other. All lines L4n-3, n a positive integer, pass through a given point A. The maximum number of points of intersection of pairs of lines from the complete set {L1, L2, . . ., L100} is

(A) 4350 (B) 4351 (C) 4900 (D) 4901 (E) 9851 Solution: (B).

One hundred lines intersect at most at 49502

991002100

=⋅

=⎟⎟⎠

⎞⎜⎜⎝

⎛ points. However, we

know that 25 lines L4, L8, . . ., L100 are parallel; hence intersections do not

exist. We are also given that the 25 lines L1, L5, . . ., L97 intersect only at point A, so that

more intersections do not exist.

300225

=⎟⎟⎠

⎞⎜⎜⎝

⎛

2991225

=−⎟⎟⎠

⎞⎜⎜⎝

⎛

The maximum number of points of intersection is 4950 – 300 – 299 = 4351. Example 28. There are six points in a plane. Sixteen triangles can be formed by connecting these points. How many lines are there that contain more than 3 of the points? Solution:

If there no three points are collinear, we can form triangles from six points. 2036

=⎟⎟⎠

⎞⎜⎜⎝

⎛

Since we are given that connecting the six points can form 16 triangles, we know that three or more points are collinear.

Since , we know that no 5 points are collinear. 10102035

26

=−=⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛

Suppose there are 4 points collinear, we have triangles, which is

true.

1642034

26

=−=⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛

Suppose there are 3 points collinear, we have

triangles, which is also true. 1642033

426

=−=⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛

26


27

So we know that either there is one line contains 4 points or three lines each containing three points. 4.4. Counting problems related to numbers Example 29: Alex wants to select two different numbers from {2, 3, 4, 5, 6, 7, 8, 9}. How many ways are there to do so such that these two numbers are not consecutive? Solution: 21. Method 1 (Listing): Case I: First integer: 2 Second integer: 4, 5, 6, 7, 8, or 9 6 ways. Case II: First integer: 3 Second integer: 5, 6, 7, 8, or 9 5 ways. Case III: First integer: 4 Second integer: 6, 7, 8, or 9 4 ways. Case IV: First integer: 5 Second integer: 7, 8, or 9 3 ways. Case V: First integer: 6 Second integer: 8 or 9 2 ways. Case VI: First integer: 7 Second integer: 9


1 way. Total 6 + 5 + 4 + 3 + 2 + 1 = 21 ways. Method 2: The number of ways to select k non-consecutive elements from n consecutive terms is

28

⎟⎟⎠

⎞

⎟⎟⎠

⎞

⎟⎟⎠

⎞

N = .

n is the total number of terms. k is the number of elements selected.

⎜⎜⎝

⎛ −−kkn

)1(

n = 9 − 2 + 1 = 8. k = 2.

N = = .

⎜⎜⎝

⎛ −−kkn

)1(

2127

)12(8

=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛ −−k

Example 30: (2006 AMC 12 A) How many non-empty subsets S of {1, 2, 3,..., 15} have the following two properties? (1) No two consecutive integers belong to S. (2) If S contains k elements, then S contains no number less than k. (A) 277 (B) 311 (C) 376 (D) 377 (E) 405 Solution: The number of ways to select k non-consecutive elements from n consecutive terms is

N = ⎜⎜ . ⎝

⎛ −−kkn

)1(

If k = 1, n = 15, then N1 = . 151 15

=⎟⎟⎠

⎞⎜⎜⎝

⎛

If k = 2, n = 14 (S contains no number less than 2), then N2 = . 782

13=⎟⎟

⎠

⎞⎜⎜⎝

⎛

If k = 3, n = 13, (S contains no number less than 3) then N3 = . 1653

11=⎟⎟

⎠

⎞⎜⎜⎝

⎛


If k = 4, n = 12, (S contains no number less than 4) then N4 = . 12649

=⎟⎟⎠

⎞⎜⎜⎝

⎛

If k = 5, n = 11, then N5 = . 2157

=⎟⎟⎠

⎞⎜⎜⎝

⎛

If k = 6, n = 10, then N6 = (not possible). ⎟⎟⎠

⎞⎜⎜⎝

⎛65

Total: 15 + 78 + 165 + 126 + 21 = 405. Note: this was the last problem of the test. Example 31: (1980 Bulgarian Mathematical Olympiad) Find the number of ways of choosing 6 among the first 49 positive integers, at least two of which are consecutive. Solution:

29

⎟⎟⎠

⎞There are ways to select 6 numbers among the first 49 positive integers. ⎜⎜

⎝

⎛6 49

There are = ways to select 6 numbers among the first 49 positive

integers such that no two of which are consecutive.

⎟⎟⎠

⎞⎜⎜⎝

⎛ −−kkn

)1(

⎟⎟⎠

⎞⎜⎜⎝

⎛6 44

The number of ways of choosing 6 among the first 49 positive integers, where at least

two of which are consecutive is

− . ⎟⎟⎠

⎞⎜⎜⎝

⎛6 49

⎟⎟⎠

⎞⎜⎜⎝

⎛6 44

Example 32. (1987 China Middle School Math Contest) Counting number n has the following property: if we take any 50 different numbers from 1, 2, 3, …, n, there always will be two numbers with the difference of 7. Among the many values of n, what is the largest value?

Solution: 98.

First, we take the 7 numbers from 1 to 7. Then we take 7 numbers again from 15 to 21 and we repeat this action until we get the last group: 85, 86, 87, 88, 89, 90, 91. Now we have 49 numbers. The difference between any two numbers in these 49 numbers is not 7.

Since the question states that we take 50 numbers, we must take 1 more number. If we take one more number (92 in our case), we will get two numbers with a difference of 7


(92 – 85). However, 92 is not the largest number since we can have any one of 93 to 98. So 98 is the largest value of n.

Note that if we leave the numbers from 92 to 98 and take 99 instead, we can go on until we take 105 so that all of these numbers will not have a difference of 7, so 99 can not be the value of n.

When n = 98, we can divided all the numbers into the following 7 subsets: },92,85,,15,8,1{ L

96,89,,19,12,5{ L

},93,86,,16,9,2{ L

}, 97,90,,20,13,6{ L

},94,87,,17,10,3{ L {}, 98,91,,21,14,7{ L

},95,88,,18,11,4 L

},

Each subset has 14 numbers and any two neighboring numbers have a difference of 7. If we take 50 numbers from these 7 sets (50 = 7 × 7 + 1) from the pigeonhole principle, we are sure that there must be one subset containing 8 numbers. There must be two consecutive numbers among these 8 numbers and there must be two numbers with a difference of 7.

So the largest value of n is 98.

Example 33: A subset of integers 1, 2,…, 100 has the property that none of its members is either the sum of other two or two times of another. What is the largest number of members such a subset can have?

Solution: 50.

All 50 odd numbers can be selected.

If we have 51 numbers, without loss of generality, we can assume that the numbers can be denoted as a1, a2, a3, .. a51, with a51 as the greatest of these numbers. Since none of these 51 members is the sum of the other two, we have the following distinct integers: a51 – a1, a51 – a2, …. a51 – a50, all of which are within 100. However, this gives us 101 total numbers. Because our subset only has 100 total numbers, among these 101 numbers, there must be two numbers that are equal. i.e. ai = a51 – ai, or 2ai = a51. This is a contradiction to the properties of our subset, so the largest number of members is 50.

Example 34: A subset of integers 1, 2,…, 10 has the property that the sum of its members is odd. How many such subsets are there?

Solution: 526.

Let A = {1, 3, 5, 7, 9} and B = {2, 4, 6, 8, 10}.

30


If we form a subset by taking out 1, 3, or 5 numbers from set A and 0, 1, 2, 3, 4, or 5 numbers from set B, the sum of the members in this subset is odd. So the number of such subsets is:

526255

45

35

25

15

05

55

35

15 9 ==⎥

⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛×⎥

⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛

Note: A subset of integers 1, 2,…, n has the property that the sum of its members is odd. The number of such subsets is 2n – 1.

A subset of integers 1, 2,…, n has the property that the sum of its members is even. The number of such subsets is 2n – 1 – 1.

31


32

PROBLEMS

Problem 1. (1986 AMC) The 120 permutations of AHSME are arranged in dictionary order, as if each were an ordinary five-letter word. The last letter of the 86th word in this list is

(A) A (B) H (C) S (D) M (E) E Problem 2. (1990 AMC) At one of George Washington’s parties, each man shook hands with everyone except his spouse, and no handshakes took place between women. If 13 married couples attended, how many handshakes were there among these 26 people?

(A) 78 (B) 185 (C) 234 (D) 321 (E) 325 Problem 3. Six cards each labeled with one of the digits 1 to 6 are put into 3 different envelopes. Each envelope will have two cards and cards 1 and 2 will be in the same envelope. How many ways are there to do so?

（A） 12 (B) 18 (C) 36 (D) 54

Problem 4. Hope High School has three elective courses for social studies and four electives for science. How many ways are there for Alex to select three electives from them this semester with at least one from each subject? (A) 30 (B) 35 (C) 42 (D) 48 Problem 5. A palindrome number is a number that is the same when written forwards or backwards. How many palindrome numbers less than 1000 are there? Problem 6. How many 3-digit even numbers can be formed by using the digit 0, 2, 3, 4, and 5? Problem 7. You have nine line segments with the lengths of 1, 2, 3, 4, 5, 6, 7, 8, and 9, respectively. How many ways are there to form a square by connecting the ends of some of these line segments? No overlapping of the line segments is allowed.


Problem 8. (1985 AMC) How many distinguishable rearrangements of the letters in CONTEST have both the vowels first? (For instance, OETCNST is one such arrangement, but OTETSNC is not.)

(A) 60 (B) 120 (C) 240 (D) 720 (E) 2520 Problem 9. In how many ways can three married couples be seated in a row if no husband and wife are to be in adjacent seats? Problem 10. A baking company produces five different cookies: Chocolate Chip Cookies, Peanut Butter Cookies, Oatmeal Cookies, Sugar Cookies, and Blueberry Cookies. If a package contains 8 cookies with at least one each kind, how many different packages are possible? Problem 11. (1985 China High School Math Contest) Find the number of nonnegative integer solutions to the equation 2x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 = 3. Problem 12. In how many ways can four men and four women be seated at a round table if no two men are to be in adjacent seats? Problem 13. In how many ways can a family of six people be seated at a round table if the youngest kid must sit between the parents? Problem 14. A gardener plants three maple trees, four oak trees, and five birch trees in a row. How many ways are there such that no two birch trees are next to one another? Problem 15. Seven identical chairs in a row are to be seated by four students. How many arrangements are there such that the only two of the three empty chairs are next to each other? Problem 16. As shown in the figure, each of six regions ABCDEF is to be assigned a color. There are 4 colors to choose from, and no adjacent regions can be the same color. How many different ways are there if each color is allowed to use more than once?

33


Problem 17. There are five regions that need to be colored by six different colors as shown in the figure. Each region can only be colored with one color. How many different ways to do the coloring?

Problem 18. (2011 AMC 10 A) Each vertex of convex pentagon ABCDE is to be assigned a color. There are 6 colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?

Problem 19. (2007 AMC 12B ) Each face of a regular tetrahedron is painted either red, white, or blue. Two colorings are considered indistinguishable if two congruent tetrahedra with those colorings can be rotated so that their appearances are identical. How many distinguishable colorings are possible?

(A) 15 (B) 18 (C) 27 (D) 54 (E) 81

Problem 20. Nine squares of a 3 × 3 board are painted using three colors: black, red and yellow with the following restrictions: (1) each square is pained with one color, (2) each color is used exactly three times, (3) each column is painted with three colors, and each row is colored with three colors. How many ways can this be done?

Problem 21. How many ways to color the 4 regions using 3 different colors, if no two neighboring regions can have the same color?

34


Problem 22. Using red, yellow, blue, and black colors to color the figure as shown. Each region is colored with one color and no adjacent region can have the same color. If each color is allowed to use more than once, how many ways are there to color? (A) 48 (B) 36 (C) 30 (D) 24 Problem 23. What is the size of the largest subset, S, of {1, 2,…, 2013} such that no pair of distinct elements of S has a sum divisible by 3? Problem 24. Line a is parallel to line b. There are 10 points on line a and 9 points on line b. At most how many points of intersection of line segments obtained by connect all these points on line a to all the points on b? Problem 25. In how many ways can two squares be selected from an 8-by-8 chessboard so that they are not in the same row or the same column? Problem 26. Alex wants to select three different numbers from {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}. How many ways are there such that no two numbers are consecutive? Problem 27. Alex wants to select four different numbers from {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}. How many ways are there such that no two numbers are consecutive? Problem 28. How many ways are there to take two different numbers from 1, 2,…, 32 such that the sum of them is divisible by 4?

Problem 29. At least how many numbers need to be removed from the list of 1, 2, 3,…, 2013 such that among the remaining numbers, any number is not a product of other two numbers?

35


36

Problem 30. A subset of integers 1, 2,…, 100 has the property that the sum of its two members is always divisible by 10. What is the largest number of members such a subset can have?

Problem 31. How many ways are there to take 7 numbers from 1 to 12 such that none of the chosen numbers is twice another? Problem 32. How many ways are there to take two different numbers from 1, 2,…, 30 such that the product of them is divisible by 7? Problem 33. How many ways are there to take three different numbers from 1, 2,…, 30 such that the sum of them is divisible by 3? Problem 34. What is the size of the largest subset, S, of {1, 2,…, 2013} such that no pair of distinct elements of S has a sum divisible by 3?


SOLUTIONS:

Problem 1: Solution: (E). The first 24 = 4! words begin with A, the next 24 begin with E and the next 24 begin with H. So the 86th begins with M, and it is the 86 – 72 = 14th such word. The first six words that begin with M begin with MA and the nest six with ME. So the desired word begins with MH and it is the second such word. The first word that begins with MH is MHAES, the second is MHASE. Thus E is the letter we seek. Problem 2: Solution: 234. If Alex shakes hand with Bob or Bob shakes hand with Alex, their interaction only counts as one time, so the order does not matter and we may use combinations to solve this problem.

We have 26 people, so at most we can have handshakes. Among these handshakes,

there are handshakes between women, and 13 handshakes between spouses.

Therefore the final answer is − − 13 = 234.

⎟⎟⎠

⎞⎜⎜⎝

⎛226

⎟⎟⎠

⎞⎜⎜⎝

⎛213

⎟⎟⎠

⎞⎜⎜⎝

⎛226

⎟⎟⎠

⎞⎜⎜⎝

⎛213

Problem 3: Solution: (B). We deal with special situation first. We have 3 ways to put cards 1 and 2 into the same

envelope. After that, we have ways to select two cards from the four remaining

cards and put them into one envelope. The two remaining cards will go in the last envelope.

624

=⎟⎟⎠

⎞⎜⎜⎝

⎛

By the product rule, we have 3 × 6 = 18 ways. Problem 4: Solution: 30. Case I: Alex can select 1 social studies course and 2 science courses. By the product rule,

1824

13

=⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟⎟⎠

⎞⎜⎜⎝

⎛.

Case II: Alex can select 2 social studies courses and 1 science course. By the product

rule, . 1214

23

=⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟⎟⎠

⎞⎜⎜⎝

⎛

By the sum rule, we have 18 + 12 = 30 ways.

37


Method 2: 3 3 37 3 4 30C C C− − = .

Problem 5: Solution: 109. Our job is to count palindrome numbers. We have the following three cases. In each case, we finish the job of counting palindrome numbers. Case I: E1: to count one-digit palindrome numbers. There are 10 one-digit palindromes: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. We finished E1 in n1= 9 ways. Case II: E2 to count two-digit palindrome numbers. There are 9 two-digit palindromes: 11, 22, 33, 44, 55, 66, 77, 88, 99. We finished E2 in n2 = 9 ways. Case III: E3 to count three-digit palindrome numbers. There are 90 3-digit palindromes: 101, 111, 121, ...999. We finished E3 in n3= 90 ways. N = n1+ n2 + n3 = 10 + 9 + 90 = 109. Problem 6: Solution: 30. The job is to form even numbers. We have the following two cases. In both cases, we finish the job of forming even numbers.

Case I: When 0 is in the ones position, we have four digits left. By )!(

!),(rn

nrnP−

= , we

have 12)!24(

!4)2,4( =−

=P such numbers.

Case II: When 0 is not in the ones position, by )!(

!),(rn

nrnP−

= , we have ways to

select the units digit (only digits 2 and 4 available), ways to select the hundred digit (0 is excluded, one of the two digits 2 or 4 is also excluded since one of them has been used already as the units digit), ways for the middle digit. By the product rule, we get:

)1,2(P

)1,3(P

)1,3(P18)1,3()1,3()1,2( =×× PPP such numbers.

38


By the sum rule, we get the final answer 12 + 18 = 30. Problem 7: Solution: 9.

We see that 124454)9321( <=÷++++ L , so the length of the side of the square is less

than 12. Case I: The side length is 11. We have one way: 9 + 2 = 8 + 3 = 7 + 4 = 6 + 5 Case II: The side length is 10. We have one way: 9 + 1 = 8 + 2 = 7 + 3 = 6 + 4 Case III: The side length is 9. We have 5 ways: 9 = 8 + 1 = 7 + 2 = 6 + 3 = 5 + 4 Case IV: The side length is 8. We have 1 way: 8 = 7 + 1 = 6 + 2 = 5 + 3 Case V: The side length is 7. We have 1 way: 7 = 6 + 1 = 5 + 2 = 4 + 3 Case VI: The side length is 6 or less. We have no ways. In total, we have 1 + 1 + 5 + 1 + 1 = 9 different ways. Problem 8: Solution: (B). There are two ways to order the vowels. By (3.1), we have the following ways to rearrange the rest of the letters C, N, T, T, S:

60!1!1!1!2

!5!!!

!

21

=×××

=⋅⋅⋅

=rnnn

nN .

By the product rule, we have the answer: 2 × 60 = 120. Problem 9: Solution: 240.

39


Let the three couples be (A, A), (B, B), (C, C). If A is seated first, we have the following 5 arrangements:

If we switch the genders for A, B, and C, we have the factors 2× 2 ×2. If A is followed by C, we get another factor of 2. The number of arrangements is 5 × 2 × 2 × 2 × 2 = 80. Similarly, B or C can also be seated first, so the final answer will be 3 × 80 = 240. Problem 10: Solution: 35. By (3.5), we have x1 + x2 + ⋅⋅⋅ + xr = n ⇒ x1 + x2 + x3 + x4 + x5 = 8

The answer is . 3547

1518

11

=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛−−

=⎟⎟⎠

⎞⎜⎜⎝

⎛−−

=rn

N

Problem 11: Solution: 174. We focus on 2x1 since this term is special. We know that x1 ≥ 0, so we have the following two cases: Case I: x1 = 0. The given equation becomes: x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 = 3. The number of nonnegative integer solutions is given by (3.6):

1653

193

1=⎟⎟

⎠

⎞⎜⎜⎝

⎛ −+=⎟⎟

⎠

⎞⎜⎜⎝

⎛ −+n

rn

Case II: x1 = 1. The given equation becomes: x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 = 1. The number of nonnegative integer solutions is given by (3.6):

40


91

191

1=⎟⎟

⎠

⎞⎜⎜⎝

⎛ −+=⎟⎟

⎠

⎞⎜⎜⎝

⎛ −+n

rn

We do not have case III since x1 is less than 2. The answer is 165 + 9 = 174. Problem 12: Solution: 144. We seat four women first. There are (4 – 1)! = 3! ways to sit them. After the ladies are seated, we have 4! ways to seat four men in the small rectangles as shown in the figure below. 4! × 3! = 144.

Problem 13: Solution: 12. We link two parents and the youngest kid together to form a unit. There are (4 – 1)! ways to seat them at the table. The result must be multiplied by 2 since we can switch the positions of the two parents, giving us the answer (4 – 1)! × 2 = 12 ways. Problem 14: Solution: 33868800. We plant the non-birch trees first. There are 7! ways to do so. There are eight spaces for

three birch trees to choose and there are 5! ways for them to rearrange themselves. ⎟⎟⎠

⎞⎜⎜⎝

⎛58

By the product rule, the answer is 7! × × 5! = 33868800. ⎟⎟⎠

⎞⎜⎜⎝

⎛58

Problem 15: Solution: 480. We tie the two empty chairs next to each other together and treat them as on unit, namely A. the other empty chair is B.

41


We have 4! = 24 ways to sit four students. There are ways to insert A and B.

and we multiply the result by 2 because A and B can switch their positions.

1025

=⎟⎟⎠

⎞⎜⎜⎝

⎛

By the product rule, the answer will be 24 × 10 × 2 = 480. Problem 16: Solution: 732. Case I: A, C, and E have the same color. The number of ways to color: 4 × 3 × 3 × 3 = 108 Case II: A, C, and E have two different colors. The number of ways to color: 3 × 4 × 3 × 3 × 2 × 2 = 432 . Case III: A, C, and E have the three different colors. The number of ways to color: P(4,3) × 2 × 2 × 2 = 192. There are a total of 108 + 432 + 192 = 732 ways to color. Problem 17: Solution: 1560. We have 6 ways to color region A first. Then we color region B (we have 5 ways), and next region C (4 ways). For the color of region D, we have two cases: Case I: Different color from B. We have 3 ways to color D (because we have three colors left) and 3 ways to color E (either from 2 colors left or the same color as C). Case II: Same color as B. We have 1 way to color D and 4 ways to color E. The answer will be 6 × 5 × 4 × (3 × 3 + 1 × 4) = 1560 ways.

42


Problem 18: Solution: 3120. Method 1: Case I: 5 colors are used:

720!556

=×⎟⎟⎠

⎞⎜⎜⎝

⎛

Case II: 4 colors are used:

18005!31446

=××××⎟⎟⎠

⎞⎜⎜⎝

⎛

We consider the vertex A first, and then we color point B and point A with the same color. We multiply by 5 because we have five vertices (5 sub cases).

Case III: 3 colors are used: ⎜⎜ 6005112336

=×××××⎟⎟⎠

⎞

⎝

⎛

We consider the vertex A first, and then we color points A and B with the same color. Finally, we multiply the result by 5 since we have 5 vertices. No other cases exist. So the final answer is 720 + 1800 + 600 = 3120.

Method 2: There is a formula for this problem which can be derived by the recursion method. An = (m – 1)[(m – 1)(n-1) + (–1)n], where n is the number of points and m is the number of colors. n = 5 and m = 6. An = (m – 1)[(m – 1)(n-1) + (–1)n] = (6 – 1)[(6 – 1)(5 – 1) + (–1)5] = 3120. Problem 19: Solution: 15. Let r, w, and b be the number of red, white, and blue faces, respectively. Case 1: One color is used. rrrr, wwww, bbbb.

43


We have 3 ways to color. Case 2: Two colors are used. rrww, rrbb, wwbb; rrrb, rrrw; wwwr, wwwb, bbbr, bbbw. We have 9 ways to color. Case 3: Three colors are used. rrwb, wwrb, bbrw. R We have 3 ways to color. In total we have 3 + 9 + 3 = 15 ways. Problem 20: Solution: 12. We have 3 ways to color square 8. Let us say we use the color black.

In the region that contains the squares 5, 9, 2, and 7, we have two ways to use the black color (color 5, 2 black or 7, 9 black). After that, we have exactly two ways to color other squares. By the product rule, we have 3 × 2 × 2 = 12 ways. Problem 21: Solution: 18. If regions 1 and 3 are painted the same color: Regions 1 2 3 4 Ways to color 3 × 2 × 1 × 2 If regions 1 and 3 are painted different colors: Regions 1 2 3 4 Ways to color 3 × 1 × 2 × 1 Total ways 12 + 6 = 18. Problem 22: (A) 48 Solution: We have two cases: Case I: Region A and region B have the same color. We have 4 ways to color region A and 1 way to color region B.

44


Then we can color region C in 3 ways and region D in 2 ways. That is 4 × 3 × 2 = 24 ways. Case II: Region A and region B have different colors. We have 4 ways to color region A, 3 ways to color region B, 2 ways to color region C, and 1 way to color region D. This gives us 4 × 3 × 2 × 1 = 24 ways. Therefore, there are a total of 24 + 24 = 48 ways to color the given figure.

Problem 23: Solution: 672. Let A = {3, 6, 9, …, 2013}, B = {1, 4, 7,…, 2011}, and C = {2, 5, 8,…, 2012}. We can at most select one element from A to add to B or C to form S. Since B and C each have 671 elements, S has at most 671 + 1 = 672 elements. Problem 24: Solution: 1620. We need two points on line a and two points on line b in in order to form 1 point of intersection. No three line segments will intersect at one point, so the maximum number of points of intersection is

162029

210

=⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟⎟⎠

⎞⎜⎜⎝

⎛.

Problem 25: Solution: 1568. Consider the coordinate system (a, b), which designates a square to be in row a, column b. In order to ensure that two squares are in different rows and different columns, both

their coordinates must be different. There are ways to choose 2 different row

coordinates and ways to choose 2 different column coordinates. Given the row

coordinates and column coordinates, there are 2 different ways to pair them into two ordered pairs. Therefore, the answer is 2 × 28 × 28 = 1568.

2828

=⎟⎟⎠

⎞⎜⎜⎝

⎛

2828

=⎟⎟⎠

⎞⎜⎜⎝

⎛

Problem 26: Solution: 220.

45


46

⎟⎟⎠

⎞

⎟⎟⎠

⎞

N = ⎜⎜ = .

Problem 27: Solution: 330.

⎝

⎛ −−kkn

)1(

220312

3 )13(14

=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛ −−

N= ⎜⎜ = .

⎝

⎛ −−kkn

)1(

330411

4 )14(14

=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛ −−

Problem 28: Solution: 120. Let A = {4, 8, 12, 16, 20, 24, 28, 32}, B = {1, 5, 9, 13, 17, 21, 25, 29}, and C = {2, 6, 10, 14, 18, 22, 26, 30}, and D = {3, 7, 11, 15, 19, 23, 27, 31}. If we take two numbers from A, or two numbers from C, the sum of them must also be divisible by 3. If we take one number from B and one number from D, the sum of them must be divisible by 3.

So the number of ways is: . 12018

18

28

2 =⎟⎟⎠

⎞⎜⎜⎝

⎛⋅⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛⋅

Problem 29: Solution: 43. We know that 44 × 45 = 1980 and 45 × 46 = 2070. So after we remove 43 numbers (2, 3,…, 44), the remaining numbers will satisfy the given condition.

On the other hand, if we only remove 42 numbers from the list, there is at least one set with all three numbers remaining.

{k, 89 – k, k(89 – k)}, k = 2, 3, …, 44.

So one number is a product of other two numbers.

Therefore we need to remove at least 43 numbers.

Problem 30: Solution: 10. If the sum of any two members is divisible by 10, both of them must be a multiple of 10, or both must have a remainder of 5 when divided by 10. We can select two numbers in the form of 10k + 5 (k = 0, 1, 2, …, 9) or 10k (k = 1，2，…，10). However, we cannot take the numbers at the same time from both groups, so we can take at most 10 numbers.


Problem 31: Solution: 47. First we classify the numbers in the following way: 1, 2, 4, 8， 3, 6, 12， 5, 10， 7， 9， 11. We can select at most 2 numbers each from (1, 2, 4, 8) and (3, 6, 12). We have three ways to take two numbers from (1, 2, 4, 8): {1, 4}, {1, 8}, {2,8}, and one way to take two numbers from (3, 6, 12): {3,12}. The other 3 numbers are selected from the remaining numbers. We also need to pay attention to these 3 numbers if we selected one of（5, 10). The number of ways to take 7 numbers is 3 × 1 ×（1 + 2 × 3）= 21. Next, we can take two numbers from (1, 2, 4, 8) and one number from (3, 6, 12). There are 3 × 3 × 2 = 18 total ways to do so.

Finally, we can take one number from (1, 2, 4, 8), and 2 numbers from (3, 6, 12). There are 4 × 1 × 2 = 8 total ways to do so.

Summing these values gives us the answer: 21 + 18 + 8 = 47. Problem 32: Solution: 110. Let A = {7, 14, 21 28} and B = {1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 13, 15, 16, 17, 18, 19, 20, 22, 23, 24, 25, 26, 27, 29, 30}. If we take two numbers from A, or we take one number from A and one number from B, the product will be divisible by 7.


14

24

=⎟⎟⎠

⎞⎜⎜⎝

⎛⋅⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛

Problem 33: Solution: 1030. Let A = {3, 6, 9, 12, 15, 18, 21, 24, 27, 30}, B = {1, 4, 7, 10, 13, 16, 19, 22, 25, 28}, and C = {2, 5, 8, 11, 14, 17, 20, 23, 26, 29}. If we take one number from A, one number from B, and one number from C, the sum of them must be divisible by 3.

47


48

1030

If we take three numbers from A, or three numbers from B, or three numbers from C, the sum of them must also be divisible by 3.


3110

110

110

=⎟⎟⎠

⎞⎜⎜⎝

⎛⋅+⎟⎟

⎠

⎞⎜⎜⎝

⎛⋅⎟⎟⎠

⎞⎜⎜⎝

⎛⋅⎟⎟⎠

⎞⎜⎜⎝

⎛

Problem 34: Solution: 672. Let A = {3, 6, 9, …, 2013}, B = {1, 4, 7,…, 2011}, and C = {2, 5, 8,…, 2012}. At most we can select one element from A to add to B or C to form S. Since B and C each has 671 elements, S has at most 671 + 1 = 672 elements.

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