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Permutations & Combinations: Selected Exercises

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.

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Permutations & Combinations: Selected Exercises

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  1. Permutations & Combinations: Selected Exercises

  2. 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).

  3. 10 There are 6different candidates for governor. In how many different orders can the names of the candidates be printed on a ballot?

  4. 10 Solution The # 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).

  5. 20 (a) How many bit strings of length 10 have exactly30s?

  6. 20 (a) Solution The bit string has 10 positions: 1, 2, …, 10. A bit string with exactly 30s 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.

  7. 20 (b) How many bit strings of length 10 have more 0s than 1s?

  8. 20 (b) Solution 1 We can decompose this problem into disjoint sub-problems, and count each sub-problem: • 60s and 41s: C(10, 6) = C(10, 4)= 10.9.8.7 / 4.3.2 = 210 • 70s and 31s: C(10, 7) = C(10, 3)= 10 .9 .8 / 3 .2 = 120 • 80s and 21s: C(10, 8) = C(10, 2)= 10 .9 / 2 = 45 • 90s and 11: C(10, 9) = C(10, 1)= 10 • 100s and 01s : 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) = 210 + 120 + 45 + 10 + 1 = 386.

  9. 20 (b) Solution 1 Is the following analysis right? • Pick the positions of 60s: C(10, 6) = C(10, 4) • Fill in the other 4 positions: 24 C(10, 4) 24 = 3,360  386. What is wrong?

  10. 20 (b) Solution 2 • There is a 1-to-1 correspondence between • strings with more 0s than 1s • strings with more 1s than 0s • Strategy: • 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.

  11. 20 (c) How many bit strings of length 10 have ≥ 71s?

  12. 20 (c) Solution We can decompose this problem into disjoint sub-problems, and count each sub-problem: • 71s and 30s: C(10, 7) = C(10, 3)= 10 .9 .8 / 3 .2 = 120 • 81s and 20s : C(10, 8) = C(10, 2) = 10 .9 / 2 = 45 • 91s and 10: C(10, 9) = C(10, 1) = 10 • 101s and 00s : C(10, 10) = C(10, 0) = 1 The answer thus is C(10, 3) + C(10, 2) + C(10, 1) + C(10, 0) = 120 + 45 + 10 + 1 = 176.

  13. 20 (d) How many bit strings of length 10 have ≥ 31s?

  14. 20 (d) Solution We 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): • 01s and 100s: C(10, 0) = 1 • 11 and 90s: C(10, 1) = 10 • 21s and 80s: C(10, 2) = 45 The answer thus is 210 – (1 + 10 + 45) = 1024 – 56 = 968.

  15. 30 (a) There are 7 women & 9 men. How many ways are there to select a committee of 5 members, with at least1 woman? Note: In such problems, it is customary and implicit to take individuals as distinct.

  16. 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?

  17. 30 (a) Solution Decompose the problem into disjoint sub-problems: • The committee has 1 woman: • Pick the woman: C(7, 1) = 7 • Pick the men: C(9, 4) = 9 .8 .7 .6 / 4 .3 .2 = 126 • The committee has 2 women: • Pick the women: C(7, 2) = 7 .6 / 2 = 21 • Pick the men: C(9, 3) = 9 .8 .7 / 3 .2 = 84 • The committee has 3 women: C(7, 3) .C(9, 2) = 35 .36 • The committee has 4 women: C(7, 4) .C(9, 1) = 35 .9 • The committee has 5 women: C(7, 5) .C(9, 0) = 21 .1 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) = 7 .126 + 21 .84 + 35 .36 + 35 .9 + 21 .1 = 4,242.

  18. 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.

  19. 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?

  20. 30 (b) Solution Subtract “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.

  21. 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? 1 6 6 5 A 2 B 1 equivalent 5 4 3 2 3 4

  22. 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: • Fix person1 at the head of the table: 1 • Arrange the other 5 people at the table: 5!

  23. 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) = k Equivalently, • Pick the captain: C(n,1) = n • Pick the remainder of the team: C(n-1,k-1)

  24. 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 . 1 Give an argument why, in general, each factor in the denominator divides some factor in the numerator.

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