No clue here on this one. Ill give it a shot and then we can all laugh and see what I did wrong.

r of N chips are red. Divide the chips into 3 groups of sizes n1,n2,andn3. where n1+n2+n3=N. Generalize the hypergrometric distribution to find the probability that the first group contains r1 red chips, the second group contains r2 red chips, and the 3rd group has r3 red chips. where r1+r2+r3=r

((r1Ck)(wCn1-k))/(NCn1))+((r2Ck)(wCn2-k)/(NCn2))+((r3Ck)(wCn3-k)/(NCn3))

I think this does it.

C(r,r1)C(N-r,n1-r1)/C(N,n1) *
C(r-r1,r2)C(N-r-n1+r1,n2-r2)/C(N-n1,n2) *
C(r-r1-r2, r3)C(N-r-n1+r1–n2+r2,n3-r3)/C(N-n1-n2,n3)

The first term is for Group 1, selecting r1 red out of r red and n1-r1 black out of N-r black divided by the number of combinations of n1 things out of N.

The second term for Group 2 is the same idea after removing the Group 1 r1 reds from the original r reds and the Group 1 n1-r1 blacks from N-r original blacks

Similarly the third term accounts for the sampling done for the first two groups.

Some simplification of the third term is possible but this form shows the symmetry -- assuming it's right, of course
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On rechecking, the third term can sure be simplified. It has to be 1. If you select r1 in group1 and r2 in group 2, then the third group has to have r3.

Note a nice check on the hypergeometric is the following

If you have C(A, a)C(B, b)C(C, c) for the numerator, then the denominator must be C(A+B+C, a+b+c).

Remember to include cases where a, b or c may be zero. For example ignoring cards in your hand, the probability a flop contains 2 clubs and 1 heart is

C(13,2)*C(13,1)*C(13,0)*C(13,0)/C(52,3)