How many different bridge hands have exactly $2$ six-card suits?
I think the answer is :
$C(13,6)$$C(4,1)$$C(13,6)$$C(4,1)$$C(26,1)$
First I choose $6$ numbers from $13$ numbers, and choose $1$ suit for one $6$ card suit.
Since there are $2$ six-card suits, I need to multiply the same thing again. And the remaining $1$ card can be chosen from $26$ cards.
Does that sound reasonable?
Best Answer
In the game of bridge, we have thirteen card hands. The phrase "exactly two six-card suits" implies exactly what it says, that there is a suit with six cards, a second different suit with six cards, and a third suit different than both previous with one card.
For example: $$\begin{array}~ \heartsuit&\heartsuit&\heartsuit&\heartsuit&\heartsuit&\heartsuit&\spadesuit&\spadesuit&\spadesuit&\spadesuit&\spadesuit&\spadesuit&\diamondsuit\\ A&2&3&4&5&6&A&2&3&4&5&6&A\end{array}$$
Break it up via multiplication principle.
Thus, there are:
$$\binom{4}{2}\binom{2}{1}\binom{13}{6}\binom{13}{6}\binom{13}{1}$$ different bridge card hands satisfying these conditions.
(note: You could have opted to use $\binom{26}{1}$ to represent choosing the one card that wasn't one of the two suits with six-cards. $\binom{26}{1}=\binom{2}{1}\binom{13}{1}$)
(note further: If you used $\binom{4}{1}$ for the first suit, and then $\binom{3}{1}$ for the second suit, you have double-counted. The sequence of choices "choose hearts, choose spades, choose diamonds, choose A23456, choose A23456, choose A" yields the same result as "choose spades, choose hearts, choose diamonds, choose A23456, choose A23456, choose A", thus we must pick both suits simultaneously and use $\binom{4}{2}$ instead)
$$\begin{array}~ \heartsuit&\heartsuit&\heartsuit&\heartsuit&\heartsuit&\heartsuit&\spadesuit&\spadesuit&\spadesuit&\spadesuit&\spadesuit&\spadesuit&\diamondsuit\\ A&2&3&4&5&6&A&2&3&4&5&6&A\end{array}$$
vs
$$\begin{array}~ \spadesuit&\spadesuit&\spadesuit&\spadesuit&\spadesuit&\spadesuit&\heartsuit&\heartsuit&\heartsuit&\heartsuit&\heartsuit&\heartsuit&\diamondsuit\\ A&2&3&4&5&6&A&2&3&4&5&6&A\end{array}$$
See the related poker question: Probability of getting two pair in poker