Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 12 - Counting and Probability - Section 12.2 Permutations and Combinations - 12.2 Assess Your Understanding - Page 875: 29


$C(4, 3)=4$ The list of possible combinations: $123,124,134, 234$

Work Step by Step

If we want to choose $r$ elements out of $n$ disregarding the order, not allowing repetition, we can do this in $C(n, r)=\frac{n!}{(n-r)!r!}$ ways. Hence, $C(4, 3)=\dfrac{4!}{(4-3)!3!}=\dfrac{4!}{1!3!}=4$ The list of all possible combinations: $123,124,134, 234$
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