The 3-digit number can be represented by 100A+10B+C.
More tens than ones: B>C.
Fewer hundreds than tens: A<B so B>A.
We only know that the tens digit is bigger than the ones and the hundreds.
The smallest B can be is 2, A<2, C<2, so because it's 3 digits A can't be zero, so C=0 and A=1 or C=A=1.
The smallest number is 120. The next possibility is 121.
If B=3, A<3, C<3 so A can be 1 or 2 and C can be 0, 1 or 2:
130, 131, 132, 230, 231, 232.
If B=4, C and A < 4, so A can be 1, 2 or 3 and C can be 0, 1, 2, 3:
140, 141, 142, 143, 240, 241, 242, 243, 340, 341, 342, 343.
...
If B=9, A can be 1, 2, 3, 4, 5, 6, 7, 8 and C can be 0, 1, 2, 3, 4, 5, 6, 7, 8:
190, 191, 192, 193, 194, 195, 196, 197, 198, ..., 890, 891, ..., 898.
So there are many solutions (2+6+12+20+30+42+56+72=240 solutions).