Permutations & Combinations: Selected Exercises

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Permutations & Combinations: Selected Exercises. 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. - PowerPoint PPT Presentation

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  • Permutations & Combinations: Selected Exercises

  • PreliminariesDenote 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).

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

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

  • 20 (a)How many bit strings of length 10 have exactly 3 0s?

  • 20 (a) SolutionThe bit string has 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, the other positions take 1s. There is 1 way to do that.The answer thus is C(10, 3) = 10 . 9 . 8 / 3 . 2 . 1 = 120.

  • 20 (b)How many bit strings of length 10 have more 0s than 1s?

  • 20 (b) Solution 1We can decompose this problem into disjoint sub-problems, and count each sub-problem:6 0s and 4 1s: C(10, 6) = C(10, 4) = 10.9.8.7 / 4.3.2 = 2107 0s and 3 1s: C(10, 7) = C(10, 3) = 10 . 9 . 8 / 3 . 2 = 1208 0s and 2 1s: C(10, 8) = C(10, 2) = 10 . 9 / 2 = 459 0s and 1 1: C(10, 9) = C(10, 1) = 1010 0s and 0 1s : C(10, 10) = C(10, 0) = 1The answer thus is C(10, 4) + C(10, 3) + C(10, 2) + C(10, 1) + C(10, 0) = 210 + 120 + 45 + 10 + 1 = 386.

  • 20 (b) Solution 1Is the following analysis right?Pick the positions of 6 0s: C(10, 6) = C(10, 4)Fill in the other 4 positions: 24

    C(10, 4) 24 = 3,360 386.What is wrong?

  • 20 (b) Solution 2There is a 1-to-1 correspondence between strings with more 0s than 1sstrings with more 1s than 0sStrategy:C(10, 5) = the # of strings with an equal # of 1s & 0s.210 C(10, 5) = the # with an unequal # of 1s & 0s.(210 C(10, 5) ) / 2 = the # with more 0s than 1s.C(10, 5) = 10.9.8.7.6 / 5.4.3.2.1 = 252(1024 252)/2 = 386.

  • 20 (c)How many bit strings of length 10 have 7 1s?

  • 20 (c) SolutionWe can decompose this problem into disjoint sub-problems, and count each sub-problem:7 1s and 3 0s: C(10, 7) = C(10, 3) = 10 . 9 . 8 / 3 . 2 = 1208 1s and 2 0s : C(10, 8) = C(10, 2) = 10 . 9 / 2 = 459 1s and 1 0: C(10, 9) = C(10, 1) = 1010 1s and 0 0s : C(10, 10) = C(10, 0) = 1The answer thus is C(10, 3) + C(10, 2) + C(10, 1) + C(10, 0) = 120 + 45 + 10 + 1 = 176.

  • 20 (d)How many bit strings of length 10 have 3 1s?

  • 20 (d) SolutionWe can decompose this problem into disjoint sub-problems, and count each sub-problem.In this case, it is easier to count the number of 10-bit strings w/o the property & subtract from the # of 10-bit strings (210):0 1s and 10 0s: C(10, 0) = 11 1 and 9 0s: C(10, 1) = 10 2 1s and 8 0s: C(10, 2) = 45The answer thus is 210 (1 + 10 + 45) = 1024 56 = 968.

  • 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?Note: In such problems, it is customary and implicit to take individuals as distinct.

  • 30 (a)Consider using the product rule:Pick 1 woman: C(7,1).Pick 4 people 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?

  • 30 (a) SolutionDecompose the problem into disjoint sub-problems:The committee has 1 woman:Pick the woman: C(7, 1) = 7Pick the men: C(9, 4) = 9 . 8 . 7 . 6 / 4 . 3 . 2 = 126The committee has 2 women:Pick the women: C(7, 2) = 7 . 6 / 2 = 21Pick the men: C(9, 3) = 9 . 8 . 7 / 3 . 2 = 84The committee has 3 women: C(7, 3) . C(9, 2) = 35 . 36The committee has 4 women: C(7, 4) . C(9, 1) = 35 . 9The committee has 5 women: C(7, 5) . C(9, 0) = 21 . 1The 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) = 7 . 126 + 21 . 84 + 35 . 36 + 35 . 9 + 21 . 1 = 4,242.

  • 30 (a) More Elegant SolutionThe 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.

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

  • 30 (b) SolutionSubtract bad committees from all 5-committees:The # of all 5-committees: C(16, 5)The # of 5-committees w/o women: C(9, 5)The # of 5-committees w/o men: C(7, 5)The answer: C(16, 5) C(9, 5) C(7, 5)= 4,368 126 21 = 4,221.

  • 40How 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?A156234B645123equivalent

  • 40 SolutionIf 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:Fix person1 at the head of the table: 1Arrange the other 5 people at the table: 5!

  • 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?Pick the k players: C(n,k)Pick the captain: C(k,1) = kEquivalently,Pick the captain: C(n,1) = nPick the remainder of the team: C(n-1,k-1)

  • Computing C(n,k)C(n,k)k = n C(n-1,k-1) C(n,k) = n/k C(n-1,k-1). Apply the above recursively, with a base case of C(n,1) = n:C(n,k) = n(n-1) . . . (n - k +1) /k!For example,C(1000, 4) = 1000 . 999 . 998 . 997 / 4 . 3 . 2 . 1Give an argument why, in general, each factor in the denominator divides some factor in the numerator.

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