Permutations & Combinations: Selected Exercises

Copyright © Peter Cappello2 Preliminaries Denote the # of arrangements of some k elements of a set of n elements as P( n, k ). Use the product rule to derive a formula for P( n, k ). Let C( n, k ) be the # of subsets of k elements drawn from a set of n elements. Use the product rule to derive a formula for C( n, k ) in terms of P( n, k ) & P( k, k ).

Copyright © Peter Cappello3 10 There are 6 different candidates for governor. In how many different orders can the names of the candidates be printed on a ballot?

Copyright © Peter Cappello4 10 Solution The # of different orders that the candidate names can be printed on a ballot is described by the following procedure: 1.Pick the candidate that appears on top (6) 2.Pick the candidate that appears below that (5) 3.Pick the candidate that appears below that (4) 4.Pick the candidate that appears below that (3) 5.Pick the candidate that appears below that (2) 6.Pick the candidate that appears below that (1) The composite number is 6! = = 720. This also is known as P(6,6).

Copyright © Peter Cappello5 20 (a) How many bit strings of length 10 have exactly 3 0s?

Copyright © Peter Cappello6 20 (a) Solution The bit strings have 10 positions: 1, 2, …, 10. A bit string with exactly 3 0s can be described as a 3-subset of the numbers 1, 2, …, 10. These are the bit positions where the 0s go. There are C(10, 3) such 3-subsets. For each such 3-subset, all other positions take 1s. There is 1 way to do that. The answer thus is C(10, 3) = / = 120.

Copyright © Peter Cappello7 20 (b) How many bit strings of length 10 have more 0s than 1s?

Copyright © Peter Cappello8 20 (b) Solution 1 Decompose this problem into disjoint sub-problems; count each sub-problem: 1.6 0s & 4 1s: C(10, 6) = C(10, 4) = / = s & 3 1s: C(10, 7) = C(10, 3) = / 3. 2 = s & 2 1s: C(10, 8) = C(10, 2) = / 2 = s & 1 1: C(10, 9) = C(10, 1) = s & 0 1s : C(10, 10) = C(10, 0) = 1 The answer thus is C(10, 4) + C(10, 3) + C(10, 2) + C(10, 1) + C(10, 0) = = 386.

Copyright © Peter Cappello9 20 (b) Solution 1 Is the following analysis right? 1.Pick the positions of 6 0s: C(10, 6) = C(10, 4) 2.Fill in the other 4 positions: 2 4 C(10, 4) 2 4 = 3,360  386. What is wrong?

Copyright © Peter Cappello10 20 (b) Solution 2 1.There is a 1-to-1 correspondence between strings with more 0s than 1s strings with more 1s than 0s 2.Strategy: 1.C(10, 5) = the # of strings with an equal # of 1s & 0s – C(10, 5) = the # with an unequal # of 1s & 0s. 3.(2 10 – C(10, 5) ) / 2 = the # with more 0s than 1s. C(10, 5) = / = 252 (1024 – 252)/2 = 386.

Copyright © Peter Cappello11 20 (c) How many bit strings of length 10 have ≥ 7 1s?

Copyright © Peter Cappello12 20 (c) Solution Decompose this problem into disjoint sub-problems, and count each sub-problem: 1.7 1s & 3 0s: C(10, 7) = C(10, 3) = / 3. 2 = s & 2 0s : C(10, 8) = C(10, 2) = / 2 = s & 1 0: C(10, 9) = C(10, 1) = s & 0 0s : C(10, 10) = C(10, 0) = 1 The answer thus is C(10, 3) + C(10, 2) + C(10, 1) + C(10, 0) = = 176.

Copyright © Peter Cappello13 20 (d) How many bit strings of length 10 have ≥ 3 1s?

Copyright © Peter Cappello14 20 (d) Solution Decompose this problem into disjoint sub-problems, & count each sub- problem. It is easier to: 1.count the number of 10-bit strings w/o the property 1.0 1s & 10 0s: C(10, 0) = & 9 0s: C(10, 1) = s & 8 0s: C(10, 2) = subtract from the # of 10-bit strings (2 10 ): The answer thus is 2 10 – ( ) = 1024 – 56 = 968.

Copyright © Peter Cappello15 30 (a) There are 7 women & 9 men. How many ways are there to select a committee of 5 members, with at least 1 woman? (In such problems, it is customary & implicit to take individuals as distinct.)

Copyright © Peter Cappello16 30 (a) Consider using the product rule: 1.Pick 1 woman: C(7,1). 2.Pick 4 members from the remaining 6 women & 9 men: C(15,4). Is the answer: C(7,1) C(15,4)? Given a committee of men & women, can you identify the stage at which each woman was chosen?

Copyright © Peter Cappello17 30 (a) Solution Decompose the problem into disjoint sub-problems: 1.The committee has 1 woman: 1.Pick the woman: C(7, 1) = 7 2.Pick the men: C(9, 4) = / = The committee has 2 women: 1.Pick the women: C(7, 2) = 7. 6 / 2 = 21 2.Pick the men: C(9, 3) = / 3. 2 = 84 3.The committee has 3 women: C(7, 3). C(9, 2) = The committee has 4 women: C(7, 4). C(9, 1) = The committee has 5 women: C(7, 5). C(9, 0) = The answer is C(7, 1)C(9, 4) + C(7, 2)C(9, 3) + C(7, 3)C(9, 2) + C(7, 4)C(9, 1) + C(7,5)C(9, 0) = = 4,242.

Copyright © Peter Cappello18 30 (a) More Elegant Solution The set of all committees with 5 members is the universe. Its size is C(7 + 9, 5). Subtract all committees w/o women: C(9, 5). The answer is C(16, 5) – C(9, 5) = 4,368 – 126 = 4,242.

Copyright © Peter Cappello19 30 (b) There are 7 women & 9 men. How many ways are there to select a committee of 5 members, with ≥ 1 woman & ≥ 1 man?

Copyright © Peter Cappello20 30 (b) Solution Subtract “bad” committees from all 5-committees: 1.The # of all 5-committees: C(16, 5) 2.The # of 5-committees w/o women: C(9, 5) 3.The # of 5-committees w/o men: C(7, 5) The answer: C(16, 5) – C(9, 5) – C(7, 5)

Copyright © Peter Cappello21 40 How many ways are there to seat 6 people around a circular table, where 2 seatings, A & B, are equivalent if A is a rotation of B? A B equivalent

Copyright © Peter Cappello22 40 Solution If the people sat in a line the answer is 6! If we drag the line seating into a circle, 6 rotations (permutations) of that “line seating” are equivalent. The answer is 6!/6 = 5! The equivalence relation has 5! equivalence classes, each with 6 elements. Alternatively: 1.Fix person 1 at the head of the table: 1 2.Arrange the other 5 people at the table: 5!

Copyright © Peter Cappello23 Computing C(n,k) How many ways are there to select a team of k players from a set of n players, with a particular player named as captain? 1.Pick the k players: C(n, k) 2.Pick the captain: C(k, 1) = k Equivalently, 1.Pick the captain: C(n, 1) = n 2.Pick the remainder of the team: C(n-1, k-1)

Copyright © Peter Cappello24 Computing C( n, k ) C( n, k )k = n C( n - 1,k - 1 )  C( n, k ) = (n / k) C( n - 1, k - 1 ), for for n ≥ k > 1 Note: n/k may not be an integer. Apply the above recursively: C( n, k ) = n(n - 1)... (n - k +1) / k!, for k ≥ 1. For example, C(1000, 4) = / Why does each factor in the denominator divide some factor in the numerator?