In shuffIing a pack of playing cards, four are accidently dropped. the probability that missing cards should be one from each suit, is

A
$\frac{1}{256}$
B
$\frac{1}{270725}$
C
$\frac{2197}{20825}$
D
None of these

Solution:

Total case $=^{52} C_{4}$
Favorable cases $= {(^{13} C_{1})}^{4}$
so, probability

$\begin{array}{l} = \frac{{(^{13} C_{1})}^{4}}{52_{C_{4}}} \\ = \frac{13 \times 13 \times 13 \times 13 \times 1 \times 2 \times 3 \times 4}{52 \times 51 \times 50 \times 49} = \frac{2197}{20825} \end{array}$

Solution:

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