Chapter 10 Counting Methods and Probability - 10.2 Use Combinations and the Binomial Theorem - 10.2 Exercises - Skill Practice - Page 694: 15

$778320$

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 $n_1=4,r_1=1$ for the ace and $n_2=52-4=48,r_2=4$ for the non-ace cards. Hence the answer: $_{4}C_1\cdot_{48}C_4=\frac{4!}{3!1!}\frac{48!}{44!4!}=4\cdot194580=778320$