HW 8 Answer Key

SECTION 4.3 46. The largest binomial coefficient occurs for r = ⌊n/2⌋. To prove this, first note that by symmetry(C(n,r) = C(n,n-r)) it is enough to show that for r < ⌊n/2⌋, C(n, r + 1) > C(n, r). Now we have C(n, r + 1) = \frac{n!}{(r+1)!·(n−r−1)!} = \frac{n!·(n−r)}{(r+1)·r!·(n−r)!} = \frac{n−r}{r+1} · C(n, r). Thus it is sufficient to show that n − r > r + 1. This is equivalent to r < n/2 − 1/2. But this follows from our original assumption that r ≤ ⌊n/2⌋ − 1.

57. The number of paths of length n of the type described equals the number of strings of length n which is 2^{n}. A path of length n of the type described must end at a point that has n as the sum of it’s coordinates, say (n − k, k) for some k between 0 and n inclusive.The number of such paths ending at (n-k,k) equals C(n − k + k, k) = C(n, k). Hence ∑_{k=0}^{n} C(n, k) = 2^{n}

SECTION 4.4 6. There are 16 cards that qualify as being an ace or a heart, so the answer is 16/52 = 4/13 ≈ 0.31. We could also compute this from Theorem 2 as 4/52 + 13/52 − 1/52.

8. \frac{C(51,4)}{C(52,5)} = \frac{5}{52} There are C(52,5) possible poker hands. There are C(51,4) possible hands with the ace of hearts. The ace takes up one of those cards, then pick 4 more from the remaining deck.

12. C(4, 1)C(48, 4)/C(52, 5)

There are 4 ways to specify the ace. One the ace is chosen for the hand, there are C(48,4) ways to choose nonaces for the remaining four cards. Therefore there are 4C(48, 4)/C(52, 5) ≈ 0.30

14. 1317888/2598960 = 2112/4165 ≈ 0.51 There are C(52, 5) = 2, 598, 960 different hands. We need to count the number of hands that have 5 different kinds/ranks. There are C(13,5) ways to choose the kinds. For each card, there are then 4 ways to choose the suit for each of those cards. Therefore there are C(13, 5) · 4^{5} = 1, 317, 888 ways to choose the hand.

16. 5148/2598960 = 33/16660 ≈ 0.0020 There are C(52, 5) = 2, 598, 960 different hands. C(4, 1)·C(13, 5) = 5148 are flushes, since we can specify a flush by choosing a suit and then choosing 5 cards from that suit.

18. There are clearly only 10.4 = 40 straight flushes, since all we get to specify for a straight flush is the starting kind in the straight(anything from ace upto 10) and the suit. Therefore the answer is 40/C(52, 5) = 40/2598960 = 1/64974.

20. 4/C(52, 5) = 4/2598960 = 1/649740.

There are 4 royal flushes, one in each suit.

24. In each case, if the numbers are chosen from integers from 1 to n, then there are C(n, 6) possible entries, only one of which is the winning one, so the answer is 1/C(n, 6) a. 1/C(30, 6) = 1/593775 ≈ 1.7 10^{−6} b. 1/C(36, 6) = 1/1947792 ≈ 5.1 10^{−7} c. 1/C(42, 6) = 1/5245786 ≈ 1.9 10^{−7} d. 1/C(48, 6) = 1/12271512 ≈ 8.1 10^{−8}

32. It is more likely with two dice. 21/216 = 0.097 vs. 5/36 = 0.139 There are 6^{2} = 36 equally possible outcomes of the roll of two dice. There are 5 ways to get a total of 8 when two dice are rolled: (6,2), (5,3), (4,4), (3,5), (2,6). There are 6^{3} = 216 equally possible outcomes of the roll of three dice. There are 21 ways to get a total of 8 when three dice are rolled: The first die could turn out to be a 6, giving rise to the 1 triple (6,1,1). The first die could be a 5, giving rise to the 2 triples (5,2,1) and (5,1,2). Continuing in this way, we see that there are 3 triples giving a total of 8 when the first die shows a 4, 4 triples when it shows 3, 5 triples when it shows a 2, and 6 triples when it shows a 1 (namely (1,6,1), (1,5,2), (1,4,3), (1,3,4), (1,2,5), and (1,1, 6)) Therefore the are 1 + 2 + 3 + 4 + 5 + 6 = 21 possible outcomes giving a total of 8.