Question

In how many ways can a pack of 52 cards be equally among 4 players in order?

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

Answer 1

The correct option is D

$\frac{52!}{(13!{)}^4}$

Here order of group is important, then the numbers of ways in which 52 different cards can be divided equally into 4 players is permutations combinations
$(\frac{52!}{4!(13!{)}^4})\times 4!=\frac{52!}{(13!{)}^4}$
$Alternativemethod:$ Each player will get 13 cards. Now first player can be given 13 cards out of 52 cards in ${}^{52}{C}_{1}3$ ways. Second player can be given 13 cards out of remaining 39 cards (i.e. 52-13=39) in ${}^{39}{C}_{13}$ ways. Third player can be given 13 cards out of remaining 26 cards (1.e.. 39-13=26) in ${}^{26}{C}_{13}$ ways and fourth player can be given 13 cards out of remaining 13 cards (i.e. 26-13=13) in ${}^{13}{C}_{13}$ ways.
Hence required number of ways
=${}^{52}{C}_{13}\times {}^{39}{C}_{13}\times {}^{26}{C}_{13}\times {}^{13}{C}_{13}$
=$\frac{52!}{13!39!}\times \frac{39!}{13!26!}\times \frac{26!}{13!13!}\times 1$
=$\frac{52!}{(13!{)}^4}$