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. $\frac{52!}{3!(17!{)}^3}$
• B. $\frac{52!}{(17!{)}^3}$
• C. $\frac{52!}{(17!{)}^4}$
• D. $\frac{52!}{3!(17!)}$

Answer 1

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

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) in
$\frac{52!}{3!(17!{)}^3}$ ways
Hence, the required number of ways
$=\frac{52!}{1!51!}\times \frac{51!}{3!(17!{)}^3}$