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Permutations and Combinations

Permutations and Combinations. Discrete Structures, Fall 2011. Permutation vs Combination. Permutations. Combinations. Ordering of elements from a set Sequence does matter 1 2 3 is not the same as 3 2 1. Collection of element from a set Order does not matter

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Permutations and Combinations

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  1. Permutations and Combinations Discrete Structures, Fall 2011

  2. Permutation vs Combination Permutations Combinations • Ordering of elements from a set • Sequence does matter • 1 2 3 is not the same as 3 2 1 • Collection of element from a set • Order does not matter • 1 2 3 is the same as 3 2 1

  3. Permutation Example • How many ways can we select 3 students from a group of 5 students to form a line to wait to order lunch? • Students = S = { 1 2 3 4 5 } • Possible permutations would be: • 1 2 3 • 3 4 5 • 2 1 3 • 5 4 2 • Solution: notice that order matters • There are 5 ways to select the first student • Once that selection is made, there are 4 ways to select the 2nd student • Then there are 3 ways to select the 3rd student. • 5*4*3 = 60. There are sixty different permutations • Alternatively, to arrange all 5 students in a line, then there are 5 * 4 * 3 * 2 * 1 = 120 permutations.

  4. Permutation Definition • A Permutation is an ordering of the objects within a distinct set. • An arrangement of a subset of the original set is called an r-Permutation, where the “r” is the number of elements in the subset • In the previous example, we first calculated the 3-permutation and then calculated the full permutation. • Consider the set S = { 1, 2, 3 } • The arrangement 2 1 3 is a permutation of S, while 2 1 is a 2-permutation of S

  5. How many permutations of a set exist? • For a given set containing n elements, we can calculate the number of possible full permutations by considering the number of options we have at each assignment • # of permutations = (n)(n-1)(n-2)(n-3)…(2)(1) = n! • n! is called “factorial of n” and is n multiplied by every number between n and 0. • 5! = 5*4*3*2*1 • 0! = 1 (special case) • For an r-permutation, we have: • # of permutations = (n)(n-1)(n-2)(n-3)…(n-r+1) • Introduce a new notation for the r-permutation: • P(n,r) = (n)(n-1)(n-2)(n-3)…(n-r+1) • P(n,r) = n!/(n-r)! • where n is the number of elements in the set, and r is the number of elements in the permutation

  6. Example • How many different ways are there to select a first-prize winner, a second-prize winner and a third-prize winner from 100 different people who entered a contest? • This is a 3-permutation • P(100,3) = 100 * 99 *98 = 970,200

  7. Combinations • How many different committees of three students can be formed by a group of 4 students? • S = { 1 2 3 4 } • Solution: • We need to find the number of subsets with three elements. • This is the same as eliminating 1 number. • There are 4 subsets, or combinations. • This is an example of finding 3-Combinations of a set S. • The number of r distinct combinations of the set of size n is denoted C(n,r) • To compute C(n,r), use the formula n!/(r!*(n-r)!)

  8. Combination Example • How many poker hands of 5 cards can be dealt from a deck of 52 cards? • Order does not matter, so this is a combination problem. • C(52,5) = 52!/5!47! • = 52*51*50*49*48/(5*4*3*2*1) = 2,598,960

  9. Class problems: • List the permutations of {a b c} • How many different permutations are there of { a b c d e f g }? • Let S = {1 2 3 4 5} • List the 3-permutations of S • List the 3-combinations of S • Find C(5,3) • Are there more n-combinations or n-permutations of a set?

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