Question

The number of ways in which $52$ playing cards can be divided into four sets, three of them having $17$ cards each and fourth are having just one card is:

• 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: (A)

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

We have to divide $52$ cards into $4$ sets in which the $4th$ set should have only one card

So, we can divide $52$ cards as $17$ , $17$ ,$17$, $1$ (you can see it in picture )

Ways to choose $1$ card form $52$ card = $(\frac{52}{1})$

Now we have $51$ card, so no. of ways to choose $17$ cards from $51$ cards = $(\frac{51}{17})$

After this, we have $34$ cards and we have to choose $17$ cards from $34$ cards

So no. of ways to choose $17$ cards from $34$ cards = $(\frac{34}{17})$

Now we have only $17$ crads so ways to choose $17$ cards from $17$ cards = $(\frac{17}{17})$

So, total number of ways to divide $52$ card into $4$ sets in which 4th set should have one card

$=(\frac{52}{1})\times (\frac{51}{17})\times (\frac{34}{17})\times (\frac{17}{17})$

And it will be equal to $=\frac{52!}{(17!{)}^3}$

Hence, option A is correct.