How many ways are there to select a three digit number $\underline{A}\ \underline{B}\ \underline{C}$ so that $A \neq B$, $B \leq C$, and $A < C$? How many ways are there to select a three digit number $\underline{A}\ \underline{B}\ \underline{C}$ so that $A \neq B$, $B \leq C$, and $A < C$?

I found the answer as $\displaystyle\sum_{k=2}^{9} (k^2-k)=240$ by evaluating some big summations. I was wondering if anyone had a solution that doesn't require evaluating big sums?
Thanks in advance.
 A: Let us split this into two cases: $A<B$ and $B<A$.
To count the number of solutions with $A<B$ we have $0<A<B\leq C\leq 9$, so we instead pick $A-1, B-A-1, C-B$, and we note that those are all non-negative integers, and if we include $9-C$ (which is also a non-negative integer) those satisfy:
$$
(A-1)+(B-A-1)+(C-B)+(9-C)=7
$$
In fact, any solution to the equation $x_1+x_2+x_3+x_4=7$ in the non-negative integers corresponds to exactly one choice of $A,B,C$ with $0<A<B\leq C\leq 9$, so we want the number of solution in the non-negative integers to the equation
$$
x_1+x_2+x_3+x_4=7
$$
The "Stars and Bars" method (Wikipedia link, see Theorem two) tells us that there are $\binom{7+4-1}{4-1}=\binom{10}{3}=120$ solutions to the above equation in the non-negative integers.
To count the number of solutions with $B<A$ we have $0\leq B < A < C\leq 9$, so we instead pick $B, A-B-1, C-A-1$, and similarly to before we get
$$
B+(A-B-1)+(C-A-1)+(9-C)=7
$$
We get the same equation, and we again want the number of solutions in the non-negative integers, which is again $120$.
So in total we get $240$ solutions.
A: If the three digits are distinct and non-zero, then there are two ways to arrange any selection of three digits from $\{1,\cdots, 9\}$.  Thus this gives us $2\times \binom 93=168$ cases.
If $B=0$, there are $\binom 92=36$ choices.
If $B=C>0$ there are $\binom 92=36$ choices.
And we are done, with $168+36+36=240$ cases.
A: If I understood your question, you want to find how many choices of $A,B,C\in\{0,1,2,3,\dots, 9\}$ such that $A\ne B$, $B\leq C$ and $A<C$.
First, $C$ can not be $0$, otherwise we would not be able to choose $A<C$.
Let us then choose $C\in\{1,2,3,\dots, 9\}$. There are $9$ ways of choosing $C$.
We then have $(C+1)-1=C$ ways of choosing $A<C$. Finally, there are $C-1$ ways of choosing $B\ne A$ such that $B\leq C$. As a result, there are :
$$\sum_{C=1}^{9}C\cdot (C-1) =240$$
ways of choosing $A,B,C$.
