Question

In how many ways can a pack of 52 cards be divided in 4 sets, three of them having 17 cards each and fourth just 1 card?

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

First we divide 52 cards into two sets which contains 1 and 51 cards respectively is

$\frac{52!}{1!51!}$
Now 51 cards can be divided equally in three sets each contains 17 cards. Here order of sets is not important $\frac{51!}{3!(17!{)}^3}$ ways.
Hence the required number of ways
=$\frac{52!}{1!51!}\times \frac{52!}{3!(17!{)}^3}$
=$\frac{52!}{1!3!(17{)}^3}=\frac{52!}{(17!{)}^{3}3!}$

$AlternateMethod$= First set can be given 17 cards out 52 cards in ${}^{52}{C}_{17}$. Second set can be given 17 cards out of remaining 35 cards (i.e. 52-17=35) in ${}^{35}{C}_{17}$. Third set can be given 17 cards out of remaining 18 cards (i.e., 35-17=18) in ${}^{18}{C}_{17}$ and fourth set can be given 1 card out of 1 card in ${}^1{C}_1$. But the first three sets can be interchanged in 3! ways. Hence the total number of ways for the required distribution.

=${}^{52}{C}_{17}\times {}^{35}{C}_{17}\times {}^{18}{C}_{17}\times {}^1{C}_1\times \frac{1}{3!}$

=$\frac{52!}{17!35!}\times \frac{35!}{17!1!}\times \frac{18!}{17!18!}\times \frac{1}{3!}$

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