The number of 3 digit numbers of the form xyz such that $x<y$ and $z\leq y$

Solution 1:

Your case 1 needs supplementing with case 1a) $x=z$, which gives $\binom 92$ additional cases, with the larger choice $=y$ and the smaller $=x=z$ -- another $36$ cases.

Then $$\overset{\text{case 1}}{2\cdot \binom 93} + \overset{\text{case 1a}}{\binom 92} + \overset{\text{case 2}}{\binom 92} + \overset{\text{case 3}}{\binom 92} = 2\cdot 84 +3\cdot 36 = 168+108 = 276$$

giving your final calculation of $2+6-7 = 1$

Solution 2:

A trial and error method looks better here.

Note that, here $$1\le x \le 8$$ $$\max(x,z) \le y\le 9$$ $$0\le z\le 9$$

Now, for $x=1$, and for all $z$, $y$ can take $8+8+8+7+6+5+4+3+2+1=52$ values.

Again, for $x=2$, and for all $z$, $y$ can take $7+7+7+7+6+5+4+3+2+1=49$ values.

So, it can be checked that, in general for $x=r$ when $1\le r \le 8$, and for all $z$, $y$ can take $(r+2)\cdot (9-r)+(8-r)+\ldots + 1= (r+2)\cdot (9-r)+\sum_{i=1}^{(8-r)} i$ values.

The value of N $$=\sum_{r=1}^8 (r+2)\cdot (9-r)+(8-r)+\ldots + 1$$ $$=\sum_{r=1}^8 (r+2)\cdot (9-r)+\sum_{r=1}^8 \sum_{i=1}^{(8-r)} i$$ $$=\sum_{r=1}^8 (7r+18-r^2)+\sum_{r=1}^8 \frac{(8-r)(8-r+1)}{2}$$

Hope this helps you.