How many three digit numbers with increasing digits can be formed from the set $\{1, 2, 3, 4, 5, 6, 7, 8\}$? 
Suppose we pick 3 numbers $x,y,z \in \{1,2,3,4,5,6,7,8\}$ and form a 3
  digit number $xyz$ how many possible combinations numbers can we
  create such that $x < y < z$. For example $357$ would be one such
  combination.

I'm struggling with this problem here is what I have so far, we know that $x \leq 6$ otherwise there aren't enough numbers in the list to make an increasing sequence. So we have $6$ choices for the first digit. Now is where I am stuck on what to do next.
Please help.
 A: Since the three digits must be distinct, we must select three of the eight elements in the set $S = \{1, 2, 3, 4, 5, 6, 7, 8\}$.  Once we have selected these digits, there is only one way to order them so that the hundreds digit is less than the tens digit, which, in turn, is less than the units digit.  Thus, the number of ways we can construct a three-digit number $100h + 10t + u$ using the elements in set $S$ in which $h < t < u$ is equal to the number of ways we can select a subset of three elements from a set of eight elements.
A: Choose three digits from $12345678$. So ${8 \choose 3} = \frac{8!}{5!3!} = 56$
A: Here is another way to see it, without (explicit) combinatorics:
Fix the middle number to be $k$, $k=2..7$.  You have $k-1$ choices for $x$ and $8-k$ choices for $z$. This gives
$$
\sum_{k=2}^7 (k-1)(8-k)=56.
$$
A: Here are two ways to proceed.


*

*How many ways are there to choose any three numbers from the set. Not all of these are admissible. But for any such trio, there is exactly one way to rearrange their order so that $x < y < z$. How many different trios when, after being reordered, give the same sequence? For instance, $321$ and $132$ both rearrange to $123$.

*First place $z$. How many ways are there to do so? (Ans: $z$ can be one of $3,4,5,6,7,8$.) Then place $y$ (the number of choices depends on $z$), and then place $x$.
I find the first to be nicer.
