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 - Chapter Test - Page 889: 8

Answer

$54,264$

Work Step by Step

If we want to choose $k$ elements out of $n$ disregarding the order (because the order of colors does not matter), not allowing repetition, we can do this with combinations. The combination formula is: $C(n,k)=\frac{n!}{(n-k)!k!}$ ways Here, the order does not matter and hence we use combinations: $\require{cancel}C(21,6)=\frac{21!}{(21-6)!6!}=\frac{21\cdot20\cdot19\cdot18\cdot17\cdot16\cdot15!}{15!6!}=\frac{21\cdot20\cdot19\cdot18\cdot17\cdot16\cdot\cancel{15!}}{\cancel{15!}6!}=\frac{21\cdot20\cdot19\cdot18\cdot17\cdot16}{6!}=54264$
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