A B + C
0 5 (2^5) => 5C0* (2^5) = 32
1 4 (2^4) => 5C1* (2^4) = 80
2 3 (2^3) => 5C2* (2^3) = 80
3 2 (2^2) => 5C3* (2^2) = 40
4 1 (2^1) => 5C4* (2^1) = 10
5 0 (2^0) => 5C5* (2^0) = 1
Total 243
With A>=2, 80 + 40 + 10 + 1 = 131
Explanation:
Let me explain the case where A = 2 and B + C = 3
we can chose 2 out of 5 letters in 5C2 ways and now B+C will have 3 letters which can be put in 2^3 ways