Question

In how many ways $13$ cards can be distributed among $4$ players equally from a deck of $52$ cards, such that each may have $A,K,Q,J$ of the same suit.

• A. $\frac{36!}{(9!{)}^4\times 4!}$
• B. $\frac{36!}{(9!{)}^4}$
• C. $\frac{52!}{(9!{)}^4}$
• D. $\frac{52!}{(9!{)}^4\times 4!}$

Answer 1

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

Given: $4$ players and each may have $A,K,Q,J$ of the same suit.
Now, Number of ways of selecting $A,K,Q,J$ of the same suit $=1$

Number of ways to selecting the other remaining $9$ cards for each player are $=\frac{36!}{(9!{)}^4\times 4!}$

Number of ways to distribute these $4$ sets $4!$

$\therefore$ Total number of ways in which cards are distributed is
$=1\times \frac{36!}{(9!{)}^4\times 4!}\times 4!=\frac{36!}{(9!{)}^4}$