## Algebra 2 (1st Edition)

$2490624$
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. We have $2$ possibilities; either we have no queen or we have $1$ queen. In the first case we have $n=48$ and $r=5$. Hence the answer: $_{48}C_5=\frac{48!}{43!5!}=1712304$ possibilities. In the second case we have $n_1=4,r_1=1$ for the queen and $n_2=48,r_2=4$ for the other $4$ cards, thus the number of possibilities: $_{4}C_1\cdot_{48}C_4=\frac{4!}{3!1!}\frac{48!}{44!4!}=4\cdot194580=778320$. Thus the total number of possibilities: $778320+1712304=2490624$