Question

The number of ways in which $52$ cards can be divided among $4$ players so that each may have $13$ is

• A. $\frac{52!}{(13!{)}^4}$
• B. $\frac{52!}{4!{}^1{}^3}$
• C. $\frac{52!}{4!(13!{)}^4}$
• D. $\frac{52!}{13!4{}^1{}^3}$

Answer 1

The correct option is A $\frac{52!}{(13!{)}^4}$

$1^{st}$ person ${=}^{52}{C}_{13}$
$2^{nd}$ person ${=}^{39}{C}_{13}$
$3^{rd}$ person ${=}^{26}{C}_{13}$
$4^{th}$ person ${=}^{13}{C}_{13}$
Total = ${}^{52}{C}_{13}$ x ${}^{39}{C}_{13}$ x ${}^{26}{C}_{13}$ x ${}^{13}{C}_{13}$
$=\frac{52!}{13!39!}\times \frac{39!}{13!26!}\times \frac{26!}{13!13!}\times \frac{13!}{13!0!}$
Total $=\frac{52!}{13!13!13!13!}$ $=\frac{52!}{(13!{)}^4}$