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Applied Combinatorics, 4th Ed. Alan Tucker

Applied Combinatorics, 4th Ed. Alan Tucker. Section 5.2 Simple Arrangements and Selections Prepared by Patrick Asaba and Patrick Scanlon. Definitions. Permutation of n distinct objects is an arrangement, or ordering, of the n objects.

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Applied Combinatorics, 4th Ed. Alan Tucker

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  1. Applied Combinatorics, 4th Ed.Alan Tucker Section 5.2 Simple Arrangements and Selections Prepared by Patrick Asaba and Patrick Scanlon Tucker Sec. 5.2

  2. Definitions • Permutation of n distinct objects is an arrangement, or ordering, of the n objects. • r-Permutation of n distinct objects is an arrangement using r of the n objects. We use P (n, r) to denote the number of r-permutations of a set of n objects. • r-Combination of n distinct objects is an unordered selection, or subset, of r out of the n objects. We use C (n, r) to denote the number of r-combinations of a set of n objects. Tucker Sec. 5.2

  3. Definitions • From the multiplication principle we obtain. • P (n,2) = n(n-1), • P (n,3) = n(n-1)(n-2), • P (n, n) = n(n-1)(n-2) x …x 3 x 2 x 1 In enumerating all permutations of n objects, we have n choices for the first position in the arrangement, n – 1 choices (the n – 1 remaining objects) for the second position,…, and finally one choice for the last position. Tucker Sec. 5.2

  4. Definitions • Using the notation n! = n (n – 1)(n – 2) … x 3 x 2 x 1, we have the formulas • P (n, n) = n! and and • P (n, r) = n (n – 1)(n – 2) x … x [n – (r – 1)] = This formula can be used to derive a formula for C (n, r). • All r-permutation can be generate by first picking r-combination of the n objects and then arranging these r objects in any order. Thus , and solving for C (n, r); • C (n, r) = = The numbers C (n, r) are frequently called binomial coefficients because of their role in the binomial expansion Tucker Sec. 5.2

  5. Examples Ranking Wizards • Ranking is simply an arrangement, or permutation, of the n things. Arrangements with Repeated Letters • r would be the letters that are not repeated • n would be the total number of letters. Tucker Sec. 5.2

  6. Example 1 – Arrangements with Repeated Letters In the word SYSTEMS how many ways can you arrange the letters so that all three S’s appear consecutively? First, we must find all possible ways for the letters that are not S (E, M, T, Y). This can be done by permutation: P(7,4) = 7!/3! = 840 different ways. Tucker Sec. 5.2

  7. Example 1 cont’d • Next, we find out where the S’s go. The “trick” to doing this is to group the S’s as one letter. So the possible letters are E, M, T, Y, and SSS. These letters can be grouped in 5!, or 120, ways. • Another approach is to use only one S and add the other two in after computing. Tucker Sec. 5.2

  8. Example 2 – Poker Probabilities • How many 5-card hands can be formed from a standard 52-card deck? • A 5-card hand is a subset of the whole 52-card deck, so there are C(52,5) = 52! / (47!5!) = 2,598,960 possible 5-card hands. Tucker Sec. 5.2

  9. Example 2 cont’d • What is the probability that a random 5-card hand from a standard deck will result in a flush (all cards of the same suit)? • First, find all subsets of 5-cards within the same suit: C(13,5) = 13! / (5!8!) = 1287 possible flushes for one suit. • Multiply that result by 4 (for the 4 suits in a deck): 1287 x 4 = 5148 possible flushes. Tucker Sec. 5.2

  10. Example 2 cont’d • To find the probability of getting a flush randomly, take the number of possible flushes and divide by the total number of possible hands: 5148 / 2,598,960 = 0.00198, or roughly 0.2% Tucker Sec. 5.2

  11. Example 2 cont’d • What is the probability of getting 3, but not 4, aces in a 5-card hand? • First, you must pick 3 of the four aces, done in C(4,3) = 4 ways. • Next, fill your hand with any 2 of the 48 other cards in the deck: C(48,2) = 1128 ways. • Multiply the results to get 1128 x 4 = 4512 ways you can get 3 aces. Tucker Sec. 5.2

  12. Example 2 cont’d • To find the probability, divide by the total number of possible hands: 4512 / 2,598,960 = 0.00174, or roughly 0.17% Tucker Sec. 5.2

  13. Example 3 – Forming Committees • A committee of k people is to be chosen from a set of 7 women and 4 men. How many ways can you form a committee if: • The committee consists of 3 women and 2 men? • The committee consists of 4 people, and Mr. Baggins must be on it? Tucker Sec. 5.2

  14. Example 3 cont’d • In case 1, you must simply pick between two subsets, men and women. Since there are three women and 2 men, the numbers are as follows: • C(7,3) x C(4,2) = 35 x 6 = 210 different ways Tucker Sec. 5.2

  15. Example 3 cont’d • In case 2, Mr. Baggins forced entry into the committee lessens our choices for who else is on the committee. But since there were no other restrictions (no gender restrictions), we can simply choose 3 from the remaining 10: C(10,3) = 120 • An easy mistake to make is to assume gender constrictions; read the problems carefully Tucker Sec. 5.2

  16. Poker Probabilities 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 let’s suppose we want unordered 5-card hands. So it is simply Tucker Sec. 5.2

  17. 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)? Tucker Sec. 5.2

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