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?
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.