Question

The number of ways in which a pack of $52$ cards of four different suits be distributed equally among $4$ players so that each may have ace, king, queen and knave of the same suit is

• A. $\frac{4!(36!)}{(9!{)}^4}$
• B. $\frac{(36!)}{(9!{)}^4}$
• C. $\frac{2(36!)}{(9!{)}^4}$
• D. $\frac{3(36!)}{(9!{)}^4}$

Answer 1

The correct option is A $\frac{4!(36!)}{(9!{)}^4}$

There are $16(=4\times 4)$ cards of Ace, King, Queen and knave.
These $4$ sets of cards can be distributed in $4$ ways.
The remaining $36$ cards are distributed among players and each player will get

$=9=\left(\frac{36}{4}\right)$ cards
Number of ways of distributing these cards

$={}^{36}{C}_9\times {}^{36-9}{C}_9\times {}^{36-18}{C}_9\times {}^{36-27}{C}_9$
$=\frac{36!}{9!27!}\times \frac{27!}{9!8!}\times \frac{18!}{9!9!}\times \frac{9!}{9!0!}=\frac{36!}{{(9!)}^4}$
$\therefore$ Total number of ways$=4!\times \frac{36!}{{(9!)}^4}=\frac{36!4!}{{(9!)}^4}$