6.Permutation and Combination
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What is the number of ways of choosing $4$ cards from a pack of $52$ playing cards? In how many of these

four cards belong to four different suits,

A

$13^{4}$

B

$13^{4}$

C

$13^{4}$

D

$13^{4}$

Solution

There will be as many ways of choosing $4$ cards from $52$ cards as there are combinations of $52$ different things, taken $4$ at a time. Therefore

The required number of ways $=\,\,^{52} C _{4}=\frac{52 !}{4 ! 48 !}=\frac{49 \times 50 \times 51 \times 52}{2 \times 3 \times 4}$

$=270725$

There are $13$ cards in each suit.

Therefore, there are $^{13} C _{1}$ ways of choosing $1$ card from $13$ cards of diamond, $^{13} C _{1}$ ways of choosing $1$ card from $13$ cards of hearts, $^{13} C _{1}$ ways of choosing $1$ card from $13$ cards of clubs, $^{13} C _{1}$ ways of choosing $1$ card from $13$ cards of spades. Hence, by multiplication principle, the required number of ways

$=\,^{13} C_{1} \times^{13} C_{1} \times^{13} C_{1} \times^{13} C_{1}=13^{4}$

Standard 11
Mathematics

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