# Probability of Bridge hand having at most two suits

I am reviewing MIT 18.440 slides. I am wondering why the question that is worded as How many bridge hands have at most two suits represented? has a given answer of ${4\choose 2}{26 \choose 13} - 8$.

I understand that the ${4\choose 2}$ takes in account choosing the two suits that the hand will have, and ${26 \choose 13}$ indicates we're choosing 13 cards out of a limited subset of the 52 total cards.

But I'm not sure where the extra $-8$ comes from. Does it include special cases such as the 4 hands that have all 13 cards from one suit?

Possible question to use as reference: A bridge hand void in one suit

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The formula $${4\choose 2}{26\choose 13}$$ includes the hand with, say 13 diamonds, thrice. Once as an extremal case of hands with clubs and diamonds only, once among the hands with diamonds and hearts only, ditto with diamonds and spades. But that hand should be counted exactly once, so we need to subtract two from the answer. Of course, the same reasoning applies to all single suit hands, so altogether we need to subtract 8.