First slide

Combinations

Question

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

Moderate

A

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

52!
C

$\frac{52!}{17!}$
D

None of these

Solution

For the first player, distribute the cards in $^{52} C_{17}$ , ways.
Now, out of 35 cards left 17 cards can be put for second player in $^{35} C_{17}$ , ways.
Similarly, for third player put them in $^{18} C_{17}$ , ways. One card for the last player can be put in $^{1} C_{1}$ , way.
Therefore, the required number of ways for the proper distribution
$\begin{array}{l} =^{52} C_{17} \times^{35} C_{17} \times^{18} C_{17} \times^{1} C_{1} \\ = \frac{52!}{35! 17!} \times \frac{35!}{18! 17!} \times \frac{18!}{17! 1!} \times 1! = \frac{52!}{(17!)^{3}} \end{array}$

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