How many ways can 5-card hands be dealt to include exactly 3 diamonds?

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A standard deck of cards consists of 4 suits (spades,hearts,diamonds,clubs) of 13 cards each. In how many ways can 5-card hands be dealt that include 3 diamonds and 2 cards from other suits ?

My answer is : \(\displaystyle C_{13}^{3}×C_{39}^{2}\)
Is it correct ?
 
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Alfredo Dawlabany said:
A standard deck of cards consists of 4 suits (spades,hearts,diamonds,clubs) of 13 cards each. In how many ways can 5-card hands be dealt that include 3 diamonds and 2 cards from other suits ?

My answer is : \(\displaystyle C_{13}^{3}×C_{39}^{2}\)
Is it correct ?
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Correct, assuming your notation is non-standard, namely the last of the C forms listed here. I would write it as C(13,3)*C(39,2).
 
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