A positive integer with three distinct digits, when added to its rever

Let the three digit no. be ABC. Now, when you add the reverse, i.e. CBA, the middle term B will get added twice. So, the middle term of the resulting three digit no. has to be EVEN.
Now, the middle term of the resulting three digit no. when ABC and CBA are added can be 4, 6 or 8 only. Why? Coz 0 and 2 are not possible since A & C cannot add up to 0 or 2 (Note: A & C are different, so 1+1 is not valid; also neither A nor C can be 0, as reversing will give a two digit number).

Now, for middle term to be Even, and 2*B to be 4 or 6 or 8, B = 2 or 3 or 4.

For 2*B=4: (A,C) = (1,3) --> One possibility.
For 2*B=6: (A,C) = (1,5), (2,4) --> 2 possibilities
For 2*B=8: (A,C) = (1,7), (2,6), (3,5) --> 3 possibilities.

Hence, in total we have 1+2+3 = 6 possible combinations. Thus, the correct answer is E.

Hope it's clear

This question is slightly unclear.

Can we count three other combinations?

1. 420 & 024 : 444
2. 630 & 036 : 666
3. 840 & 048 : 888

Now I am confused

024 is 24 lol. I dont think its a three digit integer :-D . Mods can correct me!