in Quantitative Aptitude edited by
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Three cards are chosen at random from a pack of 52 cards. In how many ways this can be done if all the three cards are of different types?

(A) 4 × 13^3

(B) 13 × 12 × 11

(C) 53C13

(D) (3 × 13) / (12 × 11)

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1 Answer

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Best answer
3 cards of different suits = (out of 4 select 3 suit) * (choose 1 out of 13 for each 3 suits )

                                                      = $\binom{4}{3} * \binom{13}{1} *\binom{13}{1} * \binom{13}{1}$

                                                      ans = 4*13*13*13

Option A is correct
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2 Comments

First card can be anyone of 52 cards..so 52 ways.

Second card can be anyone from 39 cards ( excluding the previously chosen suit)...so 39 ways

Third card can be anyone from 26 cards ( excluding previous 2 suits)...so 26 ways

Total = 52*39*26 ways = 24* 13^3 ways I am getting..

Where am I counting wrong ? Please explain
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@chandra sai In your example, some cases are repeating.

To remove this divide by 3!.
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