[Bubu]

damn ! the site ate my answer. Anyway here it is - again.
More briefly....

If anyone knows of a better way of counting the solutions
please let me know. The trick is to note that the smallest
set making a nut hand is a set of queens. We can then
enumerate the possible boards. First ignore suits

23456789TJQ
23 78 Q
23 7 9 Q
23 7 T Q
23 7 JQ
2 4 7 9 Q
2 4 7 T Q
2 4 7 JQ
2 5 7 T Q
2 5 7 JQ
2 67 JQ

That's ten for queen set nut. The counting for a king set
and an ace set are similarly tedious. I have not done.
->N_queen_nut=10

Now to count ways to avoid flushes.

case 1.
rainbow suits which needs the one board card to
carry a repeated suit. There are 4 ways of choosing
what suit to repeat. Then there are 5 ways of choosing
what board card to carry that repeated suit. The remaing
4 distinct suits can be distributed in 4! ways amongst the
remaingin board cards ->4*5*4!

case 2.
3 suits on board so 2 cards carry distinct repeated suits.
Choose the first repeated suit in 4 ways. Choose the boards
cards carrying this suit in 5 choose 2 ways (10 ways). The second repeated suit can be chosen in 3 ways. The board cards carrying this suit can then be chosen in 3 choose 2 ways (3 ways). The remaining card can be of any of the two remaing suits.->4*10*3*3*2

other cases
2 sutis on board leads to possible flushes. so that's it !

=>non-flush_combos=4*5*4!+4*10*3*2

The answer is :

(N_queen_nut+N_king_nut+N_ace_nut)*non-flush combos


Hope this is right. Please if anyone has a better way
of counting let me know.


edit:
What the hell ! How do align the columns this ? I put a nice grid to help with the counting...