Explanation Principle used: Subraction principle: Let A be a set and let U be a larger set containing A. Let A ¯ = U \ A = { x ∈ U : x ∉ A } be the complement of A in U . Then, the number | A | of objects in A is given by the rule | A | = | U | − | A ¯ | . Calculation: The deck of 52 cards is changed to a deck of 40 cards. To choose 5 cards from a deck of 40 cards, find the number of combinations as ( 40 5 ) = 658 , 008 . (i) To choose a poker hand that contains full house in which three cards of one rank and two cards of a different rank, consider the steps as follows: (a) Choose the rank with three cards. The number of ways is 10 as the only ranks from 1 to 10. (b) Choose the suit for the each card. For three same rank cards, the choices for suit are ( 4 3 ) = 4 . (c) Choose the rank with two cards. The number of ways is 10 − 1 = 9 as the only remaining rank is, 9. (d) Choose the suit for each card. For two same rank cards, the choices for suit are ( 4 2 ) = 6 . Thus, the number of combinations for a poker hand that contains full house is 10 ⋅ 4 ⋅ 9 ⋅ 6 = 2 , 160 . Probability is defined as all possible outcomes of an event per total outcomes. Hence, the probability becomes 2 , 160 658 , 008 ≈ 0.0033 . (ii) To choose a poker hand that contains a straight, consider the steps as follows: (a) Choose the five consecutive ranks. The number of ways is 10 as the ranks can begin from any number within 1 to 10. (b) Choose the suit for the card of each rank. For each card, the choices for suit are 4. Hence, the number of ways is obtained as 4 ⋅ 4 ⋅ 4 ⋅ 4 ⋅ 4 = 4 5 . Thus, the number of combinations for a poker hand that contains straight is 10 ⋅ 4 5 = 10 , 240 . Hence, the probability becomes 10 , 240 658 , 008 ≈ 0.0156 . (iii) To choose a poker hand that contains a straight flush in which five cards of consecutive ranks are from the same suit, consider the steps as follows: (a) Choose the five consecutive ranks. The number of ways is 10 as the ranks can begin from any number within 1 to 10. (b) Choose the suit for the five consecutive rank. The number of ways is 4 . Thus, the number of combinations for a poker hand that contains straight flush is 10 ⋅ 4 = 40 . Hence, the probability becomes 40 658 , 008 ≈ 0.000061 . (iv) To choose a poker hand that contains a exactly two pairs in which two cards of same ranks, two cards of different ranks and one additionally different rank, consider the steps as follows: (a) Choose the two ranks that occur in two pairs

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Chapter 2, Problem 59E

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Introductory Combinatorics

Ch. 2 - Prob. 1E Ch. 2 - How many orderings are there for a deck of 52...Ch. 2 - In how many ways can a poker hand (five cards) be...Ch. 2 - How many distinct positive divisors does each of...Ch. 2 - Determine the largest power of 10 that is a factor...Ch. 2 - How many integers greater than 5400 have both of...Ch. 2 - In how many ways can four men and eight women be...Ch. 2 - In how many ways can six men and six women be...Ch. 2 - In how many ways can 15 people be seated at a...Ch. 2 - A committee of five people is to be chosen from a...

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