Question

The number of ways in which $52$ cards can be divide 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. $\frac{51!}{(17!)3!}$

Answer 1

The correct option is A $\frac{52!}{(17!{)}^3}$

There are $52$ cards

$3$ groups should $17$ each cards.

$1$ groups should have 1 card.

no. of way of selection $={}^{52}{C}_{17}\times {}^{(52-17)}{C}_{17}\times {}^{52-34}{C}_{17}\times 1$

$=\frac{52!}{(52-17)!\times 17!}\times \frac{(52-17)!}{(52-34)!17!}\times \frac{(52-34)!}{(52-51)!\times 17!}$

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

$\text{no. of ways of solution}=\frac{\left(52\right)!}{(17!{)}^3}$