One ace will be selected from four aces and four cards will be selected from $(52-4)=48$ cards
If $P$ is the required number of ways then,
$P=C(4,1)\times C(48,4)$
$\;\;\;=\large\frac{4!}{1!(4-1)!}\times \frac{4!}{4!(48-4)!}$
$\;\;\;=\large\frac{4(3!)}{1!(3)!}\times \frac{48\times 47\times 46\times 45\times 44!}{4\times 3 \times 2 \times 1\times44!}$
$\;\;\;\;=4\times 2\times 47\times 46\times 45$
$\;\;\;=778320$ ways
Hence (A) is the correct answer.