There are no ordinary 5-high or 6-high flushes because they would be straight flushes. I'm going to calculate the number of ordinary flushes in each category. One of the rank combos for each category (two for aces) will be a straight combo, so it wouldn't count as an ordinary flush.
n(seven-high flushes) = suit combos * (rank combos - straight combos) = 4 suits * ((5*4*3*2/4*3*2*1)-1) seven-high flushes/suit = 16 seven-high flushes
I'll explain the computation for rank combos in a little more detail. I pretend like I've looked at the 7 and I somehow know that I have a 7-high flush. There are five ranks below a 7. Once I've taken the 7 out of my hand, the first of the remaining cards can be any of the 5 cards that would give me a chance to make a 7-high flush. The next card can be any of the four remaining cards and so on. How these cards are arranged doesn't matter, so I divide the arranged rank combos by the number of possible arrangements (4*3*2*1).
n(eight-high flushes) = 4 suits * ((6*5*4*3/4*3*2*1)-1) eight-high flushes/suit = 56 eight-high flushes
n(nine-high flushes) = 136
n(ten-high flushes) = 276
n(jack-high flushes) = 500
n(queen-high flushes) = 836
n(king-high flushes) = 1316
n(ace-high flushes) = 4 suits * ((12*11*10*9/4*3*2*1)-2) ace-high flushes/suit = 1972
There are two straight combos for aces because the ace is interpreted as the highest ranking card of a flush, but it would be smallest ranking card of a five-high straight. We already counted the 5-high straight flush as a straight-flush, so we can't count a2345 suited as an ace-high flush.
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