# Tucker, Applied Combinatorics1 SECTION 5.2 Simple Arrangements and Selections Aaron Desrochers and Ben Epstein.

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Tucker, Applied Combinatorics1 SECTION 5.2 Simple Arrangements and Selections Aaron Desrochers and Ben Epstein Slide 2 Tucker, Applied Combinatorics2 Vocabulary Permutation: an arrangement or ordering of n distinct objects. R-Permutation: an arrangement using r of n objects. P(n,r) denotes the r-permutation on n objects. In general, P(n,n) = n! Therefore, P(n,r) = n(n-1)(n-2)[n-(r-1)] = n!/(n-r)!. So, P(6,2) = 6!/(6-2)! = 6!/4! = (6)(5)(4)(3)(2)(1) / (4)(3)(2)(1) = (6)(5) = 30. NOTE: P(n,n) = n! Since (n-n)! = 0! = 1. So, P(4,4) = 4! = 24. Slide 3 Tucker, Applied Combinatorics3 Vocabulary Combination: an unordered arrangement of n objects. R-Combination: an unordered selection using r of n objects. C(n,r) denotes the r-combination of n objects. In general C(n,r) = n!/(r!(n-r)!). This is easily derived using P(n,r), since r- combinations of n objects gives the number of unordered arrangements, dividing P(n,r) by the total number of possible arrangements of n (so let r = n) gives the total number of unordered arrangements. So C(n,r) = P(n,r) / P(n,n) = n!/(r!(n-r)!). NOTE: A common notation for C(n,r) is. We say the combination of n choose r objects. nrnr Slide 4 Tucker, Applied Combinatorics4 Example 1 How many 5-card hands can be formed from a standard 52- card deck? First we must ask will the hands be ordered or unordered. In this example lets suppose we want unordered 5-card hands. So it is simply C(52,5) = 52!/[5!(52-5)!] = (52)(51)(50)(49)(48) /5!= 2,598,960 hands. Slide 5 Tucker, Applied Combinatorics5 Example 2 If a 5-card hand is chosen at random, what is the probability of obtaining a flush (all cards having the same suit)? In this case order does not matter, so we will use the notion of combinations. First, we must notice that there are 13 cards in every suit. Also, we have 4 suits to choose from, therefore (4)C(13,5) = 5148 ways to get a flush. The probability is found by dividing the number of ways of getting a flush by the total number of possible hands, which we computed in the last example. Therefore, the probability of a getting a flush is 5148/2,598,960 =.00198 or about.2%. Slide 6 Tucker, Applied Combinatorics6 Example 3 A committee of k people is to be chosen from a set of 7 women and 4 men. How many ways are there to form a committee of 5 people, including 3 women and 2 men? We want 3 of the 7 women and 2 of the 4 men. There are 7 choose 3 ways of selecting the three women. There are 4 choose 2 ways of selecting the 2 men. The product of the two gives the total number of ways. C(4,2)C(7,3) = Slide 7 Tucker, Applied Combinatorics7 Example 4 How many ways are there to form a committee of 4 people with at least two women members? The trick with this question is dealing with the at least part. This can be broken down as the sum of picking exactly 2 women + picking exactly 3 women + picking exactly 4 women. So, C(7,2)C(4,2) + C(7,3)C(4,1) + C(7,4)C(4,0) = This is equal to one Slide 8 Tucker, Applied Combinatorics8 Example 5 The chromatic polynomial P k (G) of graph G is the polynomial in k that gives the number of k-colorings of G. What is the chromatic polynomial of: A complete graph K 5 on five vertices (all vertices adjacent to each other) In a complete graph, each vertex must be a different color. Thus, P k (K 5 ) = P(k,5), since there are k possible colors for the first vertex, k-1 for the second choice, k-2 for the third choice and so on. k k-1 k-2 k-3 k-4 Slide 9 Tucker, Applied Combinatorics9 P(k,5) = k!/(k-5)!=k(k-1)(k-2)(k-3)(k-4) Example 5 This is the chromatic polynomial. Slide 10 Tucker, Applied Combinatorics10 Example 5 Continued b)Find P k ( C 4 ) where C 4 is a circuit of length 4? Let the vertices on the circuit C 4 be named a, b, c, d with edges (a,b), (b,c), (c,d) and (d,a). We break the computation of P k (C 4 ) into two cases, depending on whether or not a and c are given the same color. If a and c have the same color, there are k choices for the color of these two vertices. Then b and d each must only avoid the common color of a and c --- k-1color choices each. So the number of k-colorings of C 4 in this case is k(k-1)^2. If a and c have different colors, there are k(k-1) choices for the two different colors for a and then c. Now b and d each have k-2 color choices. So in this case the total number of k- colorings of C 4 is k(k-1)(k-2)^2. Combining the two cases, we obtain P k (C 4 ) = k(k-1)^2 + k(k-1)(k-2)^2. a c b d Slide 11 Tucker, Applied Combinatorics11 Class Exercise 1 How many different 8-digit binary sequences are there with six 1s and two 0s? Consider that we have 8 slots to fill given these two conditions. Any six slots must be 1s so C(8,6) gives the number of ways to fill any six slots. Notice that we do not have to make any choices for the zeros because they must go in the last two slots. So the solution is C(8,6) = 28. It is important to note that the solution can be arrived at by selecting the slots for the zeros. In this case C(8,2) = 28. Note also that this is the compliment of C(8,6) and that they are equal. So in general C(n,r) = C(n,n-r). Slide 12 Tucker, Applied Combinatorics12 Class Exercise 2 What is the probability that a 4-digit campus telephone number has 2 digits the same and the other two digits different? Again consider filling the slots, only in this case we need four of them. First we want to pick the slots. We need 2 slots to hold the pair of identical numbers. These slots can be chosen in C(4,2) ways. Note that we need not choose where the last two numbers will go because their positions are forced. Slide 13 Tucker, Applied Combinatorics13 Next, we need to choose the actual numbers. Certainly we have 10 digits to choose from and we want just one for the identical pair so C(10,1) gives us the number of ways to choose any digit. For the third digit we must select any digit that is not the same as the first, so we have C(9,1). And similarly for the fourth C(8,1). The question asks for the probability so we must divide the product of all of these combinations by the total number of possible ordered 4-digit phone numbers. Using the slot method, we have 10 choices for the first slot, 10 for the second, 10 for the third and 10 for the fourth. So 10^4 gives us the total number of possible ordered phone numbers. The probability is then [C(4,2)C(10,1)C(9,1)C(8,1)] / 10^4, 54/125 =.432 = 43.2%