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n! C(n, r) = r!(n – r)! Here we have n = 13 andr = 2. Substitute these values into the formula and simplify. n! C(n, r) = r!(n – r)! 13 ! C(13, 2) 2!(13 – 2)! In other words, there are x possible ways to pick 2 of the 13 denominations for the pairs.

n! C(n, r) = r!(n – r)! Here we have n = 13 andr = 2. Substitute these values into the formula and simplify. n! C(n, r) = r!(n – r)! 13 ! C(13, 2) 2!(13 – 2)! In other words, there are x possible ways to pick 2 of the 13 denominations for the pairs.

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...
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n! C(n, r) = r!(n – r)! Here we have n = 13 andr = 2. Substitute these values into the formula and simplify. n! C(n, r) r!(n – r)! 13 C(13, 2) = 2!(13 – 2)! = In other words, there are x possible ways to pick 2 of the 13 denominations for the pairs.
Transcribed Image Text:n! C(n, r) = r!(n – r)! Here we have n = 13 andr = 2. Substitute these values into the formula and simplify. n! C(n, r) r!(n – r)! 13 C(13, 2) = 2!(13 – 2)! = In other words, there are x possible ways to pick 2 of the 13 denominations for the pairs.
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