## Algebra 2 (1st Edition)

$48$
We know that if we want to select $r$ objects out of $n$ disregarding the order, we can do this in $_nC_r=\frac{n!}{r!(n-r)!}$ ways. Here we have $n=1\cdot(52-4)=48$ and $r=1$ (because the $4$ kings can be selected in only $1$ way, but the last choice is free). Hence the answer: $_{48}C_1=\frac{48!}{47!1!}=48$