Question

The number of ways of dividing $52$ cards amongst four players so that three players have $17$ cards each and the fourth player just one card, is?

• A. $\frac{52!}{{\left(17!\right)}^3\times 3!}$
• B. $52!$
• C. $\frac{52!}{17!}$
• D. None of these

Answer 1

The correct option is A

$\frac{52!}{{\left(17!\right)}^3\times 3!}$

Apply permutations and combinations to find the required number of ways

The first player can be dealt $17$ cards out of the deck of $52$ cards in ${}^{52}C_{17}$ ways.

The second player can be dealt $17$ cards out of the remaining $35$ cards in ${}^{35}C_{17}$ ways.

The third player can be dealt $17$ cards out of the remaining $18$ cards in ${}^{18}C_{17}$ ways.

There is only one way of getting the last player one card.

Hence, the required number of ways $=\frac{52!}{35!17!}\times \frac{35!}{18!17!}\times \frac{18!}{17!1!}\times \frac{1}{3!}$ ….($3!$ is included as none of those sets are distinct)

$=\frac{52!}{{\left(17!\right)}^3\times 3!}$

Hence, option $\left(A\right)$ is the correct answer.