Hi Chris,
Your answer is better than mine in that I certainly did not need to get into permutations.
My logic did lead to a correct answer I think (as verified below) but it made it unnecessarily messy and there was far too much room to make a numeric error.
However:
Your answer is not correct because you have not included all the possibilities.
I'll try again. Using your ideas.
Ways to get all 5 from the same suit = 4*13C5 (Just like Chris said)
Ways to get 4 from one suit and 1 from another = 4*13C4* 39 (you forgot this one Chris)
Ways to get 3 from one suit and 2 from another = 4*13C3* 3*13C2
Total number of ways to choose 5 cards from 52 = 52C5
nCr(52,5) = 2598960
Ways to get no more than 2 suits = 4*13C5 + 4*13C4* 39 + 4*13C3* 3*13C2
4*nCr(13,5)+4*nCr(13,4)*3*13+4*nCr(13,3)*3*nCr(13,2) = 384384
Number of ways of NOT getting 2 suits = 52C5 - 384384
nCr(52,5)-384384 = 2214576
So
P(drawing at least 3 different suits) = 2214576 / 2598960
2214576/2598960 = 0.8521008403361345 = 507/595
Which is exactly the same as I got doing it with permutations !!
WOW