I've been given not a problem, but a claim and a "proof" that I have to find a problem in.
The claim is that in a 52 deck of cards, the number of ways to select a 5 hand card with at least 3 black cards is ${26 \choose 3} \cdot {49 \choose 2}$
The "proof" is that they are selecting three cards from 26 black ones, and then picking 2 from the remaining 49 (23 black cards + 26 red cards)
Now, I know that this is wrong and that they are over-counting, but I don't understand what they are overcounting. Any help is appreciated. Thanks.
Best Answer
You're double counting because the hands with $4$ (or more) black cards are counted in more than way: say we have a hand with the black cards $b_1,b_2,b_3,b_4$. We could have selected $b_1,b_2,b_3$ among the $\binom{26}{3}$ group and $b_4$ among the "remaining cards" or $b_1,b_2,b4$ as the group of $3$ and $b_3$ as one of the remaining etc. So that hand is arrived at in at least two different ways in this counting scheme of "$3$ plus remaining". In fact we count it four times, this way, and a hand with $5$ black cards is even counted $10$ times.