Question

In how many ways the cards can be divided in $4$ sets, $3$ of them having $17$ cards each and the $4th$ will be one card?

• A. $\frac{52!}{3!\times 17{!}^3}$
• B. ${}^{52}{C}_{17}\times 3!$
• C. $52!$
• D. none of these

Answer 1

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

Total numbers of cards $=52$

Arrangements $=52!$

Three sets $⟶{(17!)}^3\times 3!$

and fourth set single card

Therefore total number of ways

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