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AMS 301 - Homework 5 Solutions

5.1.22

Split into sequential cases based on the digit. For the first digit there is no constraint, so there are3 possibile numbers to use. For each choice of the first number there are 2 choices for the second

number, as the preceding digit is disallowed. Similarly, there are 2 choices for every subsequentdigits, so there are 3(29) = 1536 different ternary sequences of length 10 without any pair ofconsecutive digits the same.

5.1.30

There are 100, 000 numbers between 1 and 100, 000. We can add in the first 0, and remove100, 000 without changing the number of 5s, so look at the 100, 000 numbers between 0 and99, 999. Write each number as a five digit number by putting in preceding zeros when neces-sary, so 14 is 00014. Look at the number of fives in each position separately. In the first digit(from the right), the digits cycle 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 as the number increases. This cycle has 10digits in it, so to cycle through 100, 000 numbers needs 10, 000 cycles. For each cycle, we haveexactly 1 five, so the first digit contributes exactly 10, 000 fives. In the second digit, the digits cy-cle 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3 . . . as the numbers increase.This cycle has 100 digits in it, so to cycle through 100, 000 numbers needs 1000 cycles. For eachcycle, we have exactly 10 fives, so the second digit contributes 10, 000 fives. Continue onwards, andnote that each digit contributes exactly 10, 000 fives, meaning there are 50, 000 fives in the numbers1 to 100, 000.

a. 4 is incorrect. I cant think of a way to legitimately think 4 is the correct answer.

b. 5 104 is the correct answer, see above.

c. 1 + 10 + 100 + 1000 is incorrect. It tries to count (incorrectly) the numbers that contain a

five, not the number of fives.

5.2.12

a. This is P(14; 3, 5, 6), that is, 14!3!5!6!

= 168, 168

b. The number of ways of splitting into groups of 7 is P(14; 7, 7). However, this considers eachunordered split twice, as the split (1, 2, 3, 4, 5, 6, 7) gives the same unordered groups as thesplit (8, 9, 10, 11, 12, 13, 14). Therefore, the correct answer is 1

2P(14; 7, 7), that is, 14!

27!7!= 1716.

5.2.24

There are 109

different 9 digit numbers. There are P(10, 9) different 9 digit numbers without arepeated digit. Therefore, there is a 10

910!

109probability of a randomly selected 9 digit number

having a repeated digit.

5.2.30

First, choose the locations of the vowels. There are C(7, 3) ways to do this. The vowels must gointo their spaces in alphabetic order, so there is only 1 way to arrange them. Then arrange theconsonants in the remaining for spaces in P(4, 4) ways. Therefore, the answer is C(7, 3)P(4, 4).

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a. 3!4! is incorrect. It finds the number of ways of arranging the vowels and the consonantsseparately.

b. 7!3!

is the correct answer (in a different form)

c. C(7, 3)4! is the correct answer (in a different form)

5.3.2

MISSISSIPPI has 11 letters. Splitting into classes, there are 4 Is, 4 Ss, 2 Ps and 1 M. Thereforethere are P(11; 4, 4, 2, 1) ways to arrange the letters.

5.3.10

a. There are C(5+31, 5) = C(7, 5) ways to distribute the apples. If each person needs a pear,we distribute 3 pears in 1 way and then distribute the remaining 3 pears with no restrictionsin C(3+ 3 1, 3) = C(5, 3) ways. Therefore, there are C(7, 5)C(5, 3) ways of distributing thefruit

b. There are 35 ways of distributing the apples. To figure out how to distribute the pears, wenote that there are 10 ways to divide 6 into 3 numbers that are at least 1. That is, 6 breaksdown into 1, 1, 4 in 3 ways, into 1, 2, 3 in 6 ways and into 2, 2, 2 in 1 way. For each of thesebreakdowns, we use the formula to count the number of ways of distributing distinct thingsin specific ways. Therefore, there are 3P(6;1, 1, 4) + 6P(6;1, 2, 3) + 1P(6;2, 2, 2) ways ofdistributing the pears so each person gets at least 1. Then, there are 3 5[3 6!

1!1!4!+ 6 6!

1!2!3!+

1 6!2!2!2!

] ways of distributing all of the fruit so that each person gets at least one pear.

5.4.2

a. Without restriction, this is just 416.

b. Choose the children to get 6 toys in C(4, 2) ways. Then, there are P(16; 6, 6, 2, 2) ways ofdistributing the toys. Therefore, there are C(4, 2)P(16;6, 6, 2, 2) ways to distribute the toyswith two children each getting 6 and two children each getting 2.

c. This is just P(16; 4, 4, 4, 4)

5.4.14

We know that there are C(n + r 1, r) nonnegative integer solutions to x1 + x2 + + xn = r.However, x1 + x2 + x3 + x4 < 100 is not in that form. Firstly, the question asks for positivesolutions, not nonnegative ones. Secondly, it has a < instead of an =. First, we convert from

positive to nonnegative by subtracting 1 from each xi. This means that we can solve an equivalentproblem of finding nonnegative solutions to y1 + y2 + y3 + y4 < 96. Now we need to change the

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5.4.30

a. Condition on the number of nickels used. If we use 1 nickel, then the question is askingfor the number of ways to make 30 cents change using two different year pennies. There areexactly 31 ways to do that (Giving the number 0, 1, . . . , or 30 1952 pennies completely definesthe way to give change). Similarly, if we use k nickels, there are 35 5k + 1 ways to make

change with the pennies. In total, there are

7k=0(35 5k + 1) ways to make change, whichis 36 + 31 + 26 + 21 + + 6 + 1 = 148.

b. In a similar way as above, condition of the number of nickels in the two cases, when a quarter isused and when no quarter is used. Then, there are

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k=0(355k+1)+

2

k=0(35255k+1) =

148 + (11 + 6 + 1) = 166 ways to make change using those coins.

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