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.
I would break this into two cases... hands which hold the ace of diamonds, and hands that don't, and then add them together.
$1{3\choose 2}{12\choose 6}{36\choose4} + {3\choose 3}{12\choose 7}{36\choose 3}$