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  • How to solve this problem- In how manyways can a pack of 52 cards be divided into 4 sets, 3 of them having 16 cards each and the fourth one having just 4 cards.

In how many ways can a pack of 52 cards be divided into 4 sets, 3 of them having 16 cards each and the fourth one having just 4 cards.

  • Option 1)

    \frac{52!}{\left ( 16! \right )^{3}\left ( 3! \right )^{2}}

  • Option 2)

    \frac{1}{4}\left [ \frac{52!}{\left ( 16! \right )^{3}\left ( 3! \right )^{2}} \right ]

  • Option 3)

    \frac{1}{4}\left [ \frac{52!}{\left ( 16 \right )^{3}\left ( 4! \right )^{2}}\right ]

  • Option 4)

    \frac{1}{2}\left [ \frac{52!}{\left ( 16! \right )^{3}\left ( 3! \right )^{2}} \right ]

 

Answers (1)

best_answer

As we learnt in 

Rule for Division into Groups -

Number of ways in which m\times n different objects can be distributed equally among n persons =\frac{(mn)! n!}{(m!)^nn!}=\frac{(mn)!}{(m!)^{n}}

-

 

 52 Cards-16,16,16,4

No. of ways =\frac{52!}{\left ( 16! \right )^{3}\left (3! \right )\left ( 4! \right )}

Since there are three groups of same number.


Option 1)

\frac{52!}{\left ( 16! \right )^{3}\left ( 3! \right )^{2}}

Incorrect

Option 2)

\frac{1}{4}\left [ \frac{52!}{\left ( 16! \right )^{3}\left ( 3! \right )^{2}} \right ]

Correct

Option 3)

\frac{1}{4}\left [ \frac{52!}{\left ( 16 \right )^{3}\left ( 4! \right )^{2}}\right ]

Incorrect

Option 4)

\frac{1}{2}\left [ \frac{52!}{\left ( 16! \right )^{3}\left ( 3! \right )^{2}} \right ]

Incorrect

Posted by

Aadil

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