Answer
4
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
