I've been having some trouble with formulating the probabilities of Straights of 7, 6, 5, 4 and 3 in a 7-Card Hand. Since the total number of hands is 52C7, which is 133784560 possibilities, I've calculated Straight of 7 to be 7C1 x (4C1)^7, choosing 1 of the straights out of the possible 7 (starting on 2, highest straight starting on an 8) and the 4C1^7 to determine the suit of each card. This comes out at about 0.09%.
Straight of 6 I've determined to be (2 x 24 x (4C1)^6) + (6 x 20 x (4C1)^6) = 688128, or 0.51%. I've determined this because there are 2 possible straights (2-3-4-5-6-7 and 9-10-J-Q-K-A) that have 7 cards that form a Straight of 7 from them, the 6 cards in the straight and the 1 card below and above them respectively. 7 x 4 for suit = 28, and 52 - 28 = 24. This applies to the other 6 Straight-of-6's, with the 6 cards in the straight and the two above and below not being allowed to be counted. 8 x 4 for suit = 32, 52 - 32 = 20. 4C1^6 again to determine suits of the cards.
However I've become a bit stumped when trying to form a Straight of 5. I've used the same concept, and come out with (2 x 28 x (4C1)^5) + (7 x 24 x (4C1^5), however this comes out as a lower number than Straight of 6 and I've just noticed that I'm missing a card in itself, and I'm not sure how to include this into the equation. Help would be awesome, and some assistance on calculating Straight of 4 / Straight of 3 with the same concept would be amazing. Thanks!