Homework4key

# Homework4key - 1 Homework 4(11.5 points due Feb 10(1 pt...

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Unformatted text preview: 1 Homework 4 (11.5 points) due Feb. 10 (1 pt.) 3.34ab A club has 14 members. a) How many ways can a governing committee of size 3 be chosen? This is without replacement because once a person is on the committee; he can’t be on it again. This is unordered because there are no individual titles to the members on the committee. N(E) = 14 3 ¡ = 14! ¢ 14 − 3 £ !3! = 14! 11!3! = 14 ∙ 13 ∙ 12 3! = 364 b) How many ways can a president, vice president, and treasurer be chosen? This is without replacement for the same reason. This is ordered because the president, vice president and treasurer are distinct positions. N(E) = (14) 3 = 14! ¢ 14 − 3 £ ! = 14! 11! = 14 · 13 · 12 = 2184 = 14 3 ¡ 3! = 14! ¢ 14 − 3 £ !3! 3! (2.5 pts.) 3.36 (abdef only) (I will post the answers to all of the parts for your personal information :) (p.107) In five-card draw, the order of the cards is not important. In addition, aces can either be high or low, that is the order is: ace, 2, 3, …, jack, queen, king, ace. Determine the number of possible hands of the specified type. a) straight flush: five cards of the same suit in sequence N(E) = (the number of ways that 1 suit can be chosen) times (the number of possible ending or starting cards of the straight). We have 14 numbers in a suit because Ace can be either high or low. ¢£ = ¤ 4 1 ¥ ¢ 14 − 5 + 1 £ = 4 ∙ 10 = 40 b) Four of a kind: {w, w, w, w, x}, where w and x are distinct denominations. N(E) = (the number of ways that 1 number can be chosen from the numbers) times (the number of ways that 4 cards can be chosen from 4 cards) times (the number of ways that 1 card can be chosen from the rest of the deck) ¢£ = ¤ 13 1 ¥ ¤ 4 4 ¥ ¤ 52 − 4 1 ¥ = 13 ∙ 1 ∙ 48 = 624 d) flush: five cards of the same suit, not all in sequence N(E) = (the number of ways that 5 cards can be chosen from the suit) times (the way that 1 suit can be chosen from all of the suits) - (the number of straight flushes (part a)) ¢£ = ¤ 13 5 ¥ ¤ 4 1 ¥ − 40 = 13! ¢ 13 − 5 £ ! 5! ∙ 4 − 40 = 13! 8! 5! ∙ 4 − 40 = 5148 − 40 = 5108 2 e) straight: five cards in sequence, not all of the same suit N(E) = (the number of possible ending cards of a straight – see part a) times (the number of ways that 1 number can be chosen from 4 numbers repeated 5 times) - (the number of straight flushes (part a)) ¡ = 10 ¢ 4 1 £ ¢ 4 1 £ ¢ 4 1 £ ¢ 4 1 £ ¢ 4 1 £ − 40 = 10 ∙ 4 5 − 40 = 10,240 − 40 = 10,200 f) three of a kind: {w, w, w, x, y}, where w, x and y are distinct denominations N(E) = (the number of ways that 1 number can be chosen from all of the numbers) times (the number of ways that 3 cards can be chosen from 4 cards) times (the number of ways that 2 cards can be chosen from the non-w numbers in the deck - unordered) times (the number of ways that 1 number can be chosen from 4 suits repeated twice)....
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## This note was uploaded on 02/20/2012 for the course STAT 311 taught by Professor Staff during the Spring '08 term at Purdue.

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Homework4key - 1 Homework 4(11.5 points due Feb 10(1 pt...

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