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.
- Choose which two suits are each represented by six cards: (we must choose both of these simultaneously since otherwise we cannot distinguish between having chosen $\spadesuit$ and then $\heartsuit$ versus $\heartsuit$ and then $\spadesuit$) (Thus there are $\binom{4}{2}$ number of choices for this step)
- Choose which suit is represented by a single card: ($\binom{2}{1}$ number of choices for this)
- Choose which numbers are used in each suit: (at this point we have already chosen which suit is which and can distinguish between them. For each of the suits with six cards there will be $\binom{13}{6}$ choices and for the suit with one card there will be $\binom{13}{1}$ choice)
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
Best Answer
By the Pigeon Hole Principle: the fifth card must be one of those four suits also.
Count ways to select: a suit to have two cards, two cards for that suit, and one card for each of the three others.
Assuming this is a standard 52 card deck (4 suits with 13 cards each and no jokers).