Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 11 - Probability and Statistics - 11-1 Permutations and Combinations - Practice and Problem-Solving Exercises - Page 678: 33

Answer

$4$

Work Step by Step

Use the formula $C(n,r)=\dfrac{n!}{r!(n-r)!}$ to obtain: $C(4,3)=\dfrac{4!}{3!(4-3)!}$ $C(4,3)=\dfrac{4!}{3!(1!)}$ $C(4,3)=\dfrac{ 4 {\times} 3 {\times} 2 {\times} 1}{(3{\times}2 {\times} 1)(1)}$ Simplify to get $C(4,3)=4$ There are $4$ possible combinations.
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