First slide

Combinations

Question

The number of ways of selecting $4$ cards of an ordinary pack of playing cards so that exactly $3$ of them
are of the same denomination is

Moderate

A

2496
B

$^{13} C_{3} \times^{4} C_{3} \times 48$
C

$^{52} C_{3} \times 48$
D

none of these

Solution

We can choose one denomination in $^{13} C_{1}$ ways, then $3$ cards of this denomination can be chosen
in $^{4} C_{3}$ ways and one remaining card can be chosen in $^{48} C_{1}$ ways. Thus, the total number of choices is
$(^{13} C_{1}) (^{4} C_{3}) (^{48} C_{1}) = 13 \times 4 \times 48 = 2496$ .

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