If the groups ARE distinct, the problem is easier: just (9 choose 3) * (6 choose 3), then you're set.

Brent@GMATPrepNow wrote:

puneetkhurana2000 wrote:Note:- This question should be out of scope for GMAT.

I think this is a 700+ level question, but I don't think it's out of scope (IMHO)

Cheers,
Brent

It's a great example of how difficulty is determined as much by obscurity as by the challenges of the problem itself. If you've done a little combinatorics, the problem is pretty easy, but enough of the GMAT test taking population HASN'T done combinatorics that the problem is assigned a high rank. (The same is true of many standard deviation questions.)

puneetkhurana2000 wrote:Note:- This question should be out of scope for GMAT.

Nah, definitely fair game. The GMAT would use smaller numbers, though, something like 4 horses for 2 carts or 6 horses for 3.