Homework Help is Here – Start Your Trial Now!
Math
Probability
The required number of ways in which poker hands are consist of two cards of one rank, two cards of another different rank, and one card of third rank are 123552 ways.

The required number of ways in which poker hands are consist of two cards of one rank, two cards of another different rank, and one card of third rank are 123552 ways.

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
See similar textbooks
icon
Related questions
Question

Why isn't the answer C(13,1)C(4,2) x C(12,1)C(4,2) x C(11,1)C(4,1) using the same logic later in the provided answer?

To calculate: The number of ways in which poker hands consist of two cards of one rank, two cards of another different rank, and one card of third rank. Answer to Problem 32E The required number of ways in which poker hands are consist of two cards of one rank, two cards of another different rank, and one card of third rank are 123552 ways. Explanation of Solution Given Information: It is given that a poker hand consists of two cards of one rank, two cards of another different rank, and one card of third rank. A poker hand consists of 5 cards selected from a standard deck of 52 cards. Formula used: The formula for total number of combinations, P(n,r) r! C(n,r) = n! r!(n-r)! = Calculation: Consider the given statement that a poker hand consists of two cards of one rank, two cards of another different rank, and one card of third rank. There are total 4 rank available with 13 card each in a standard deck of 52 cards. So, number of combinations of choose 2 cards out of 4, having 13 similar cards, 13C (4,2) = 13 ( 2!(4-2)! 4! = 13 (2) = 13 x 6 = 78 Next, again choose two cards of another different rank, so this case also has similarity with 13 cards in first case. So, total number of combination poker hand consists of two cards of one rank, two cards of another different rank, 21(13-2)! (21(4-2)! ) (214-2)!) 4! 13×12×11! (41) (241) (2121) 2!x11! 2!2! = 78 × 6 × 6 = 2808 Now, as two kind of rank cards are already chosen, there are 13 – 2 = 11 cards available to choose. Thus, the total number of ways one cards of third rank can be choosed are, 11C(4, 1) = 11 ( C (13, 2) × (C (4, 2) × C (4, 2)) = 4x3! = 11 (43) 1!×3! 11 (4) = = 44 4! 1!(4-1)! = Hence, the number ways in which poker hands two cards of one rank, two cards of another different rank, and one card of third rank. 2808 × 44 = 123552. Therefore, the required number of ways in which poker hands are consist of two cards of one rank, two cards of another different rank, and one card of third rank are 123552 ways.
Transcribed Image Text:To calculate: The number of ways in which poker hands consist of two cards of one rank, two cards of another different rank, and one card of third rank. Answer to Problem 32E The required number of ways in which poker hands are consist of two cards of one rank, two cards of another different rank, and one card of third rank are 123552 ways. Explanation of Solution Given Information: It is given that a poker hand consists of two cards of one rank, two cards of another different rank, and one card of third rank. A poker hand consists of 5 cards selected from a standard deck of 52 cards. Formula used: The formula for total number of combinations, P(n,r) r! C(n,r) = n! r!(n-r)! = Calculation: Consider the given statement that a poker hand consists of two cards of one rank, two cards of another different rank, and one card of third rank. There are total 4 rank available with 13 card each in a standard deck of 52 cards. So, number of combinations of choose 2 cards out of 4, having 13 similar cards, 13C (4,2) = 13 ( 2!(4-2)! 4! = 13 (2) = 13 x 6 = 78 Next, again choose two cards of another different rank, so this case also has similarity with 13 cards in first case. So, total number of combination poker hand consists of two cards of one rank, two cards of another different rank, 21(13-2)! (21(4-2)! ) (214-2)!) 4! 13×12×11! (41) (241) (2121) 2!x11! 2!2! = 78 × 6 × 6 = 2808 Now, as two kind of rank cards are already chosen, there are 13 – 2 = 11 cards available to choose. Thus, the total number of ways one cards of third rank can be choosed are, 11C(4, 1) = 11 ( C (13, 2) × (C (4, 2) × C (4, 2)) = 4x3! = 11 (43) 1!×3! 11 (4) = = 44 4! 1!(4-1)! = Hence, the number ways in which poker hands two cards of one rank, two cards of another different rank, and one card of third rank. 2808 × 44 = 123552. Therefore, the required number of ways in which poker hands are consist of two cards of one rank, two cards of another different rank, and one card of third rank are 123552 ways.
To calculate: The number of ways in which poker hands consist of two cards of one rank, two cards of another different rank, and one card of third rank. Answer to Problem 32E The required number of ways in which poker hands are consist of two cards of one rank, two cards of another different rank, and one card of third rank are 123552 ways. Explanation of Solution Given Information: It is given that a poker hand consists of two cards of one rank, two cards of another different rank, and one card of third rank. A poker hand consists of 5 cards selected from a standard deck of 52 cards. Formula used: The formula for total number of combinations, P(n,r) r! C(n,r) = n! r!(n-r)! = Calculation: Consider the given statement that a poker hand consists of two cards of one rank, two cards of another different rank, and one card of third rank. There are total 4 rank available with 13 card each in a standard deck of 52 cards. So, number of combinations of choose 2 cards out of 4, having 13 similar cards, 13C (4,2) = 13 ( 2!(4-2)! 4! = 13 (2) = 13 x 6 = 78 Next, again choose two cards of another different rank, so this case also has similarity with 13 cards in first case. So, total number of combination poker hand consists of two cards of one rank, two cards of another different rank, 21(13-2)! (21(4-2)! ) (214-2)!) 4! 13×12×11! (41) (241) (2121) 2!x11! 2!2! = 78 × 6 × 6 = 2808 Now, as two kind of rank cards are already chosen, there are 13 – 2 = 11 cards available to choose. Thus, the total number of ways one cards of third rank can be choosed are, 11C(4, 1) = 11 ( C (13, 2) × (C (4, 2) × C (4, 2)) = 4x3! = 11 (43) 1!×3! 11 (4) = = 44 4! 1!(4-1)! = Hence, the number ways in which poker hands two cards of one rank, two cards of another different rank, and one card of third rank. 2808 × 44 = 123552. Therefore, the required number of ways in which poker hands are consist of two cards of one rank, two cards of another different rank, and one card of third rank are 123552 ways.
Transcribed Image Text:To calculate: The number of ways in which poker hands consist of two cards of one rank, two cards of another different rank, and one card of third rank. Answer to Problem 32E The required number of ways in which poker hands are consist of two cards of one rank, two cards of another different rank, and one card of third rank are 123552 ways. Explanation of Solution Given Information: It is given that a poker hand consists of two cards of one rank, two cards of another different rank, and one card of third rank. A poker hand consists of 5 cards selected from a standard deck of 52 cards. Formula used: The formula for total number of combinations, P(n,r) r! C(n,r) = n! r!(n-r)! = Calculation: Consider the given statement that a poker hand consists of two cards of one rank, two cards of another different rank, and one card of third rank. There are total 4 rank available with 13 card each in a standard deck of 52 cards. So, number of combinations of choose 2 cards out of 4, having 13 similar cards, 13C (4,2) = 13 ( 2!(4-2)! 4! = 13 (2) = 13 x 6 = 78 Next, again choose two cards of another different rank, so this case also has similarity with 13 cards in first case. So, total number of combination poker hand consists of two cards of one rank, two cards of another different rank, 21(13-2)! (21(4-2)! ) (214-2)!) 4! 13×12×11! (41) (241) (2121) 2!x11! 2!2! = 78 × 6 × 6 = 2808 Now, as two kind of rank cards are already chosen, there are 13 – 2 = 11 cards available to choose. Thus, the total number of ways one cards of third rank can be choosed are, 11C(4, 1) = 11 ( C (13, 2) × (C (4, 2) × C (4, 2)) = 4x3! = 11 (43) 1!×3! 11 (4) = = 44 4! 1!(4-1)! = Hence, the number ways in which poker hands two cards of one rank, two cards of another different rank, and one card of third rank. 2808 × 44 = 123552. Therefore, the required number of ways in which poker hands are consist of two cards of one rank, two cards of another different rank, and one card of third rank are 123552 ways.
Expert Solution
steps

Step by step

Solved in 2 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Similar questions
Recommended textbooks for you
A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
Probability
ISBN:
9780134753119
Author:
Sheldon Ross
Publisher:
PEARSON
A First Course in Probability
A First Course in Probability
Probability
ISBN:
9780321794772
Author:
Sheldon Ross
Publisher:
PEARSON
SEE MORE TEXTBOOKS