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How many $5$ card poker hands contain at least $1$ red and $1$ black card?

I used inclusion-exclusion to calculate my answer. The number of total poker card hands are:$$52\choose 5$$I have $26$ red cards and $26$ black cards as my $A$ and $B$ set. I also need the total number of hands that have flushes as my $(A\cap B)$. So my formula is as follows:$$1red1black=U -A+B-(A\cap B)$$So:$$1red1black={52\choose 5}-26+26-\left(2\cdot \left({13\choose 5}{4\choose 1}-{10\choose 1}{4\choose 1}\right)\right)$$$$={52\choose 5}-2\cdot \left(4{13\choose 5}-40\right)$$Is this line of thinking correct?

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    $\begingroup$ I don't understand your line of thinking, but I got a different answer from you. What is $U$? Is $U$ the set of all possible hands? If $A$ is the set of all 26 red cards, why are you subtracting that set from $U$? It doesn't make sense to subtract a set of cards from a set of hands. $\endgroup$ – Ceph May 9 '16 at 19:36
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The OP calculation uses a formula incorrectly.

The correct value is $${52\choose 5}-{26\choose 5}-{26\choose 5}$$ This counts the total number of 5-card hands, less the 5-card hands that are entirely red cards, less the 5-card hands that are entirely black cards. There are no 5-card hands that are both entirely red and entirely black, so there is no inclusion-exclusion needed here.

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  • $\begingroup$ Why exactly are we subtracting the flush twice? $\endgroup$ – Jodo1992 May 11 '16 at 3:21
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    $\begingroup$ One for all red, one for all black. $\endgroup$ – true blue anil May 11 '16 at 3:37
  • $\begingroup$ I know order of operations can be finnicky when working with binomial coefficients, but would this be the same as ${52\choose 5}-2\cdot {26\choose 5}$? $\endgroup$ – Jodo1992 May 14 '16 at 19:52
  • $\begingroup$ Yes, that is equal. Pineapple minus banana minus banana equals pineapple minus two bananas. $\endgroup$ – vadim123 May 14 '16 at 20:06

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