Question

The number of ways of distributing $52$ cards among four players so that three players have $17$ cards each and the fourth player has just one card?

• A. $\frac{52!}{3!{\left(17!\right)}^3}$
• B. $52!$
• C. $\frac{52!}{17!}$
• D. $\frac{52!}{{\left(17!\right)}^2}$

Answer 1

The correct option is A

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

The explanation for the correct answer.

Find the number of ways of distributing cards.

Number of ways for the distribution of the cards

$$\begin{gathered} ={}^{52}{\mathrm{C}}_{17}\times {}^{35}{\mathrm{C}}_{17}\times {}^{18}{\mathrm{C}}_{17}\times {}^1{\mathrm{C}}_1\times \frac{1}{3!} \\ =\frac{52!}{17!\left(52-17\right)!}\times \frac{35!}{17!\left(35-17\right)!}\times \frac{18!}{17!\left(18-17\right)!}\times 1\times \frac{1}{3!}\mathbf{}\left({}^n{C}_r=\frac{n!}{r!\left(n-r\right)!},{}^n{C}_n=1\right) \\ =\frac{52!}{17!\times 35!}\times \frac{35!}{17!\times 18!}\times \frac{18!}{17!\times 1}\times 1\times \frac{1}{3!} \\ =\frac{52!}{{\left(17!\right)}^3\times 3!} \end{gathered}$$

We divide it by $3!$ because players having $17$ cards can be interchanged.

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