Question

The number of ways in which 52 cards can be divided into 4 sets, three of them having 17 cards each and the fourth one having just one card –

• A. $\frac{52!}{(17!)^{3}}$
• B. $\frac{52!}{(17!)^{3}3!}$
• C. $\frac{51!}{(17!)^{3}}$
• D. $\frac{51!}{(17!)^{3}3!}$

Answer 1

Answer: (B)

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

Answer 2

Here we have to divide 52 cards into 4 sets, three of them having 17 cards each and the fourth one having just one card. First we divide 52 cards into two groups of 1 card and 51 cards. This can be done in 52! / 1! 51! ways.

Now every group of 51 cards can be divided into 3 groups of 17 each in $\frac{51!}{(17!)^{3}3!}$

Hence the required number of ways

$\frac{52!}{1!51!}$·$\frac{51!}{(17!)^{3}3!}$=$\frac{52!}{(17!)^{3}3!}$.