Combi for Fun

Let's do casework (sometimes we can find nice patterns by doing so!):

For $A = 5$ , $B$ can be anything in $[1, 4]$ ( $B$ cannot be $0$ otherwise we will not have any value for $C$ . ) The value of $B$ then determines the possible value(s) of $C$ : We can deduce that $C$ has possible values in $[0,3]$ -- When $B=4$ , $C$ can take $4$ values, i.e. $0,\; 1,\; 2,\; 3$ ; When $B=3$ , $C$ can take $3$ values; ... etc. ...

Thus, for $A=5$ , the number of cases is $\sum_{r=1}^{4}r$ .

Similarly, we have the number of cases for $A=6,\: 7,\: 8,\: 9$ , giving the required sum

$\sum_{r=1}^{4}r+\sum_{r=1}^{5}r+\sum_{r=1}^{6}r+\sum_{r=1}^{7}r+\sum_{r=1}^{8}r= \boxed{110}$ .

For each value of A (from 5 up to 9) , B can have values 1 up to (A-1)

For each A,B pair, C can have values 0 up to (B-1) i.e. B number of different values.

So each A contributes 1+2+....+(A-1) numbers, i.e. A(A-1)/2 numbers.

For values of A from 9 to 5, it comes out:

36+28+21+15+10 = 110