Number of words with length 3 with certain restrictions In today's quiz (about combinatorics) there was a question I was not able to solve mathematically:
Find the number of words with length 3 over an alphabet $ \Sigma = \{1,2,...,10\} $ where $x \lt y \le z$. ($x$ is the first letter, and so on).
How can I solve this in an efficient way? Of course you can just count it like:
$9 \times 10 \times 10 \rightarrow$ 1 possiblity
$8 \times \{9,10\} \times \{9,10,10\} \rightarrow$ 2 + 1 possibilities
....
But how do I solve it with "combinatoric" methods?

There are two kinds of length 3 words $w = xyz$ satisfying the conditions that $x < y \le z$ and $x, y, z \in \Sigma = \{1, 2, 3, \dots, 10\}$:


*

*Those with $y = z$
These are in 1-1 correspondence with the set of all 2-element subsets of $\Sigma$, so there are $\binom {10} 2$ of them.

*Those with $y < z$
These are in 1-1 correspondence with the set of all 3-element subsets of $\Sigma$, so there are $\binom {10} 3$ of them.
So the number of strings satisfying the conditions is
$$
\begin{align}
\binom {10} 2 + \binom {10} 3 &= 45 + 120 \\
&= 165 \text{.}
\end{align}
$$
