# Six-card hand with $2$ cards of each suit - did I calculate this correctly?

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?

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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.
If we disregard the order in which the terms are listed (so $$3,2,1$$ is the same as $$2,1,3$$), I count at least $$7$$ ways to add three non-negative integers with sum $$6.$$ And you say you observe one of them with probability greater than $$\frac17.$$ So this would not be a particularly unlikely combination at that rate.
For the $$3,2,1$$ hand you presumably don’t care which suit is the one with $$3$$ cards or which has $$2$$ cards. If you do care then you already have the correct probability (and there are several other events with that exact same probability); otherwise you should multiply by the number of ways you can select different suits for the $$3$$-card, $$2$$-card, and $$1$$-card suits.
If you still have doubts you can work out the other five distributions of numbers of cards of each suit and check that the sum of all probabilities is $$1.$$