Originally Posted by
Jame I'm sorry I was actually mistaken in my answer for 2). Since we have different amounts of each suit (5,6,2), we must choose each suit individually. So instead of 4C3, its 4*3*2 (or 4!)
The rest of the answer is the same.
Example: a hand with "5 hearts, 6 spades and 2 diamonds" is different then a hand with "5 diamonds, 6 hearts and 2 spades"
(order of the suits does in fact matter here)
If we had a different breakdown of suits (like 5,5,3). We would have to choose the two suits with the 5 cards (4C2), then choose the suit with 3 cards (2C1)
The only time we would choose the suits all together is if each card had the same breakdown (e.g. (3,3,3))