Suppose you draw five cards from a deck of 52 cards. What is the probability of getting three cards (but not four cards) of one kind (all of their suits are same)?
Basically, it seems intuitively to me that $\frac{C(13,3)C(4,1)C(48,1)}{C(52,5)}$ is the correct answer, since we pick arbitrarily $5$ from $52$ cards, 3 cards of same kind from each $13$ possibilities, multiplying it with $4$ and choosing remaining card.
For the sake of completeness, 2 2 2 K 6 and 2 2 2 K K are each three cards but 2 2 2 2 K is not.
"Three of a kind" in card problems usually means three of the same value (Ace, King, Queen, etc.) rather than three of the same suit. The OP, however, has parenthetically explained that they mean three of the same suit. Given that interpretation, $C(13,3)C(4,1)$ does describe the number of ways of choosing those three cards. But it remains to choose two more cards, not just one, and they must come from one of the other three suits (since the OP explains parenthetically that there are to be exactly three, not four), so the final multiplicative factor should be $C(39,2)$, not $C(48,1)$. The probability is thus
$$C(13,3)C(4,1)C(39,2)\over C(52,5)$$
Just for completeness, if "three of kind" is given its usual meaning, the numerator for the probability is $C(13,1)C(4,3)C(48,2)$. I.e., choose the value for the three of a kind, then choose three of the four cards of that value, and finally choose two cards from the remaining cards of the other values.