Given abcd is a 4 digit number, such that a,b,c,d are distinct non zero digits & a + b + c = d

Hence lets start from the max value that d can take, d = 9, hence we have {a,b,c} = {6,2,1} or {5,3,1} or {4,3,2}

each combination of {a,b,c} can be arranged among themselves in 3! = 6 ways.

Therefore for d = 9, we have 6 * 3 = 18 ways

Similarly for d =8, we have {a,b,c} = {5,2,1} or {4,3,1}, hence 12 ways

for d = 7, we have {a,b,c} = {4,2,1}, hence 6 ways

& for d = 6, we have {a,b,c} = {3,2,1}, hence 6 ways.

we cannot take d < 6, since we don't have {a,b,c} such that they satisfy a+b+c = d

Hence Total # of 4 digit numbers required = 18 + 12 + 6 + 6 = 42

Answer E.

Thanks,

GyM

_________________

New to GMAT Club - https://gmatclub.com/forum/new-to-gmat-club-need-help-271131.html#p2098335