In a game of Texas Hold'Em you are dealt: 4 of hearts and 5 of hearts
The flop: 2-club, 6-spade, K-Diamond.
Define set S1 as 7-card hands that contain the 5 revealed cards AND contain a straight AND its turn and river are 7 and 8, regardless of order.
What is the size of S1?
Define set S2 as 7-card hands that contain the 5 revealed cards AND contain a straight AND its turn and river contain 3.
What is the size of S2?
Please help me with this
As for the first, the 5 cards are absolute, for the turn and river, there are four 7's and four 8's, so I guess S1's size is ${4 \choose 1} * {4 \choose 1} = 16$
For the second: it has to contain straight, it must contain 3 and it already has 2, 4, 5 and 6, which means that way it already has a straight. There are four 3's in the deck, so there are ${4 \choose 1}$ ways to pick up a 3. As for the 7'th card, it can be any card. How many cards are left after all this selection? $52-6 = 46$. So totally it's ${4 \choose 1} * 46 = 184$