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: 59



Work Step by Step

Use the formula of combination: $_{n}$C$_{r}$=$\frac{n!}{r!(n-r)!}$.You have $_{2}$C$_{2}$, $_{2}$C$_{1}$ and $_{2}$C$_{0}$. Solve each of them separately and add the solutions at the end. -->$_{n}$C$_{r}$=$\frac{n!}{r!(n-r)!}$ $_{2}$C$_{2}$=$\frac{2!}{2!(2-2)!}$ -simplify like terms- $_{2}$C$_{2}$=$\frac{2!}{2! (0!)}$ -write using factorial- $_{2}$C$_{2}$=$\frac{2*1}{(2*1)(1)}$ -simplify- $_{2}$C$_{2}$=1 -->$_{n}$C$_{r}$=$\frac{n!}{r!(n-r)!}$ $_{2}$C$_{1}$=$\frac{2!}{1!(2-1)!}$ -simplify like terms- $_{2}$C$_{1}$=$\frac{2!}{1! (1!)}$ -write using factorial- $_{2}$C$_{1}$=$\frac{2*1}{(1)(1)}$ -simplify- $_{2}$C$_{1}$=2 -->$_{n}$C$_{r}$=$\frac{n!}{r!(n-r)!}$ $_{2}$C$_{0}$=$\frac{2!}{0!(2-0)!}$ -simplify like terms- $_{2}$C$_{0}$=$\frac{2!}{0! (2!)}$ -write using factorial- $_{2}$C$_{0}$=$\frac{2*1}{(1)(2*1)}$ -simplify- $_{2}$C$_{0}$=1 The solutions are 1+2+1=4. So $_{2}$C$_{2}$ +$_{2}$C$_{1}$ +$_{2}$C$_{0}$=4
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