## Algebra 2 Common Core

$10$
Add $_{4}$C$_{3}$ + $_{6}$C$_{5}$.Solve each separately using the formula $_nC_r=\dfrac{n!}{r!(n-r)!}$: $_4C_3=\dfrac{4!}{(3!(4-3)!)}$ $_4C_3=\dfrac{4!}{3!(1!)}$ $_4C_3=\dfrac{4*3*2*1}{(3*2*1)(1)}$ $_4C_3=4$ $_6C_5=\dfrac{6!}{(5!(6-5)!)}$ $_6C_5=\dfrac{6!}{5!(1!)}$ $_6C_5=\dfrac{6*5*4*3*2*1}{(5*4*3*2*1)(1)}$ $_6C_5=6$ Thus, $_4C_3+_6C_54+6=10$