Question

The number of ways a pack of $52$ cards can be divided among four players in $4$ sets, three of them having $17$ cards each and the fourth one just $1$ card is :

• A. $\frac{52!}{(17!{)}^3}\times 4$
• B. $\frac{52!}{3\cdot (17!{)}^3}$
• C. $\frac{52!}{3!(17!{)}^3}$
• D. $\frac{52!}{(3!{)}^3(17!)}$

Answer 1

The correct option is A $\frac{52!}{(17!{)}^3}\times 4$

Makeing the $3$ group of $17$ and out of $4$ one can be selected in ${}^4{C}_1$ ways
and all the groups will be different because distributon is done in $3$ different player.

Hence the required number of ways $=\frac{52!}{(17!{)}^3}\times {}^4{C}_1=\frac{52!}{(17!{)}^3}\times 4$