Number of ways to get a ten-high hand? So we're drawing a hand of 5.  How many ways are there to get a 10high?
I was thinking it would be this.
For the first card, you get no choice -  it's a 10, but it can be one of 4 suits. The next cards can be anything lower than 10 (consider ace to be one), there are 36 cards less than 10, so we choose 4.
This gives an answer of:
$$1 \cdot _4C_1 \cdot _{36}C_4$$
Could someone confirm this answer?
EDIT: I should note that I'm not taking into account things like straights and flushes. Mostly, I just want to know if my logic is very flawed or not.
Also, it would be very helpful if someone could show me how to do this question minus the flushes and straights.
Thanks again!
Thanks!
 A: You can avoid pairs, three of a kinds, and four of a kinds from the beginning by first insisting that you choose distinct faces.
Step 1: Choose any 10. ($4$ ways)
Step 2: Choose four distinct faces from 2 through 9. ($\binom{8}{4}$ ways)
Step 3: Assign suits to each of the four cards chosen in Step 2 ($4^4$ ways)
The count so far is 
$$
4 \cdot \binom{8}{4} \cdot 4^4.
$$
This count almost works, but there remains the trouble of straights, flushes, and straight flushes.
The faces in a 10-high hand are already determined, but we can assign one of four suits to each card. Thus, there are $4^5$ 10-high straights.
The number of 10-high flushes could be counted as follows.
Step 1: Choose a suit. ($4$ ways)
Step 2: Choose four faces (not five, since one must be a 10) from that suit. ($\binom{8}{4}$ ways)
The revised (but still wrong) count is
$$
4 \cdot \binom{8}{4} \cdot 4^4 - 4^5 - 4 \cdot \binom{8}{4}.
$$
The problem is that we've subtracted the straight flushes twice - once because the hand is a straight and a second time because it is a flush. To counter this mistake, we need to add the number of straight flushes. Fortunately, there are only four of these, giving a final count of 
$$
4 \cdot \binom{8}{4} \cdot 4^4 - 4^5 - 4 \cdot \binom{8}{4} + 4 = 70,380.
$$
