Question

The number of ways a pack of $52$ cards be equally distributed among $4$ players in order are

• A. $\left(\frac{52!}{(13!{)}^4}\right)\times 4!$
• B. $\frac{52!}{(13!{)}^4}$
• C. $\frac{52!}{4!(13!{)}^4}$
• D. ${}^{52}{P}_{13}{\times }^{39}{P}_{13}{\times }^{26}{P}_{13}{\times }^{13}{P}_{13}$

Answer 1

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

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