Suppose you have a deck of $36$ cards - $3$ different suits, $12$ cards per suit. If you draw a $6$-card hand, what is the chance of a hand with $2$ cards of each suit ($2-2-2$)?
I would do $$\frac{\left(\binom{12}2\right)^3}{\binom{36}6} = 0.1476$$ But this seems very small. Am I doing this correctly?
For a hand of $3-2-1$ ($3$ of one suit, $2$ of another suit, $1$ of the last), I would do $$\frac{\binom{12}3 \cdot \binom{12}2 \cdot \binom{12}1}{\binom{36}6} = 0.0895$$ Again, this seems quite small. Are these correct?
Your first answer is correct. In your second answer, your answer is too small because you did not take into account from which suit three cards are drawn, from which suit two cards are drawn, and from which suit one card is drawn.
Choose the suit from which three cards will be drawn, choose three of the twelve cards of that suit, choose from which of the two remaining suits two cards will be drawn, choose two of the twelve cards of that suit. The remaining card must be drawn from the remaining suit. Choose one of the twelve cards of that suit. $$\frac{\binom{3}{1}\binom{12}{3}\binom{2}{1}\binom{12}{2}\binom{1}{1}\binom{12}{1}}{\binom{36}{6}} = \frac{3!\binom{12}{3}\binom{12}{2}\binom{12}{1}}{\binom{36}{6}}$$ where the factor of $3!$ accounts for the number of ways we can select from which suit three cards are drawn, from which suit two cards are drawn, and from which suit one card is drawn.