How many three digit numbers with increasing digits can be formed from the set $\{1, 2, 3, 4, 5, 6, 7, 8\}$?

Solution 1:

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.

Solution 2:

Choose three digits from $12345678$. So ${8 \choose 3} = \frac{8!}{5!3!} = 56$

Solution 3:

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 $$ \sum_{k=2}^7 (k-1)(8-k)=56. $$