Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 12 - Data Analysis and Probability - 12-6 Permutations and Combinations - Practice and Problem-Solving Exercises - Page 768: 65



Work Step by Step

A)Find $_{6}$C$_{4}$: $_{n}$C$_{r}$=$\frac{n!}{r!(n-r)!}$ $_{6}$C$_{4}$=$\frac{6!}{4!(6-4)!}$ -simplify like terms- $_{6}$C$_{4}$=$\frac{6!}{4! (2!)}$ -write using factorial- $_{6}$C$_{4}$=$\frac{6*5*4*3*2*1}{(4*3*2*1)(2*1)}$ -simplify- $_{6}$C$_{4}$=15 B)There are 3 more questions to be answered and 6 questions left. C)Find $_{6}$C$_{3}$ from information from B. $_{n}$C$_{r}$=$\frac{n!}{r!(n-r)!}$ $_{6}$C$_{3}$=$\frac{6!}{3!(6-3)!}$ -simplify like terms- $_{6}$C$_{3}$=$\frac{6!}{3! (3!)}$ -write using factorial- $_{6}$C$_{3}$=$\frac{6*5*4*3*2*1}{(3*2*1)(3*2*1)}$ -simplify- $_{6}$C$_{3}$=20 D.Multiply the results from A and C using the multiplication counting principle: 15$\times$20=300 The answer is 300.
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