How many unused input combinations are there in a BCD adder

How many unused input combinations are there in a BCD adder? 256 312 512 128 less than 128

Solution

the answer is B -312

let me justify it as

there ate 9 inputs making 512 possible input combinations. hence 10 valid combinations for A and 10 valid combinations for b , two valid combinations for Carry In. is equals to 10 x 10 x 2 = 200 valid combinations.

Now:

512 possible - 200 valid = 312 invalid (left out )