Answer
$10$
Work Step by Step
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, $_4C_3+_6C_54+6=10$
