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) $\binom {53} {13}$
(D) $(3 × 13) / (12 × 11)$
My approach:
We can select the first card out of 52 cards. Then the next card out of 39 cards and the next one out of 26 cards.
So, $52×39×26/3!$ which is equivalent to option (A). Is this the right approach to solve this question? Am I wrong somewhere?
Your approach is okay, though for the sake of completeness you should also mention why you divided by $3!$.
Another approach is first choosing $3$ out of $4$ types, and after that $1$ card out of each of the chosen types. That gives $\binom43\times13^3$.