Answer
$1$
Work Step by Step
Use the formula $C(n,r)=\dfrac{n!}{r!(n-r)!}$ to obtain:
$C(4,4)=\dfrac{4!}{4!(4-4)!}$
$C(4,4)=\dfrac{4!}{4!(0!)}$
$C(4,4)=\dfrac{ 4 {\times} 3 {\times} 2 {\times} 1}{( 4{\times}3{\times}2 {\times} 1)(1)}$
Simplify to get
$C(4,4)=1$
There is $1$ possible combination.