## Algebra 1: Common Core (15th Edition)

Use the formula of combination: $_{n}$C$_{r}$=$\frac{n!}{r!(n-r)!}$.You have $_{3}$C$_{3}$, $_{3}$C$_{2}$, $_{3}$C$_{1}$, and $_{3}$C$_{0}$. Solve each of them separately and add the solutions at the end. -->$_{n}$C$_{r}$=$\frac{n!}{r!(n-r)!}$ $_{3}$C$_{3}$=$\frac{3!}{3!(3-3)!}$ -simplify like terms- $_{3}$C$_{3}$=$\frac{3!}{3! (0!)}$ -write using factorial- $_{3}$C$_{3}$=$\frac{3*2*1}{(3*2*1)(1)}$ -simplify- $_{3}$C$_{3}$=1 -->$_{n}$C$_{r}$=$\frac{n!}{r!(n-r)!}$ $_{3}$C$_{2}$=$\frac{3!}{2!(3-2)!}$ -simplify like terms- $_{3}$C$_{2}$=$\frac{3!}{2! (1!)}$ -write using factorial- $_{3}$C$_{2}$=$\frac{3*2*1}{(2*1)(1)}$ -simplify- $_{3}$C$_{2}$=3 -->$_{n}$C$_{r}$=$\frac{n!}{r!(n-r)!}$ $_{3}$C$_{1}$=$\frac{3!}{1!(3-1)!}$ -simplify like terms- $_{3}$C$_{1}$=$\frac{3!}{1! (2!)}$ -write using factorial- $_{3}$C$_{1}$=$\frac{3*2*1}{(1)(2*1)}$ -simplify- $_{3}$C$_{1}$=3 -->$_{n}$C$_{r}$=$\frac{n!}{r!(n-r)!}$ $_{3}$C$_{0}$=$\frac{3!}{0!(3-0)!}$ -simplify like terms- $_{3}$C$_{0}$=$\frac{3!}{0! (3!)}$ -write using factorial- $_{3}$C$_{0}$=$\frac{3*2*1}{(1)(3*2*1)}$ -simplify- $_{3}$C$_{0}$=1 The solutions are 1+3+3+1=8. So $_{3}$C$_{3}$ + $_{3}$C$_{2}$+$_{3}$C$_{1}$ +$_{3}$C$_{0}$=8