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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

Option: 1

2496


Option: 2

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


Option: 3

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


Option: 4

none of these


Answers (1)

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

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Divya Prakash Singh

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