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

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

$\begin{array}{l} =^{52} C_{17} \times^{35} C_{17} \times^{18} C_{17} \times^{1} C_{1} \\ = \frac{52!}{35! 17!} \times \frac{35!}{18! 17!} \times \frac{18!}{17! 1!} \times 1! = \frac{52!}{(17!)^{3}} \end{array}$

Solution:

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