# How many bridge hands contain exactly three aces or exactly seven diamonds, or both?

How many bridge hands contain exactly three aces or exactly seven diamonds, or both?

I'm guessing that this problem is:

$$|A \cup B| = |A| + |B| - |A \cap B|$$

So I start off by calculating $|A|$, or the combinations of exactly $3$ aces, then I pick the rest of the bridge hand:

$$|A| = {}_4C_3 * {}_{48}C_{10}$$

Then I calculate $|B|$, or the combinations of exactly 7 diamonds, then I pick the rest of the bridge hand:

$$|B| = {}_{13}C_7 * {}_{39}C_{6}$$

I then have to calculate $|A \cap B|$, which is where I'm lost. I'm not sure how to do this. I had a few guesses.

The first involves picking $3$ of the $4$ aces, then assuming one of the aces chosen was a diamond, picking 6 of the 12 remaining diamonds, then the rest of the cards:

$$|A \cap B| = {}_{4}C_3 * {}_{12}C_{6} * {}_{36}C_{4}$$

The second involves picking $3$ of the $4$ aces, then assuming one of the aces chosen was NOT a diamond, picking $7$ of the $12$ remaining diamonds, then the rest of the cards:

$$|A \cap B| = {}_{4}C_3 * {}_{12}C_{7} * {}_{36}C_{3}$$

I thought that maybe I'd have to add both of these cases to get $|A \cap B|$, but this is not giving me the correct answer when inserting these values into:

$$|A \cap B| = |A| + |B| - |A \cap B|$$

Any help is greatly appreciated. This problem has been challenging me for a day now.

$|A∩B|$ exactly 3 aces and exactly 7 daimonds.
$1{3\choose 2}{12\choose 6}{36\choose4} + {3\choose 3}{12\choose 7}{36\choose 3}$
• Couldn't they sum to 52 if your initial term was ${1\choose 0}$? Sep 21, 2017 at 2:35