Each of the digits 1, 2, 3, 4, 5, 6 is used exactly once in forming the 3-digit integers X and Y. How many possible values of X+Y are there if|X−Y|=111?
6 C 3 ==20 combinations. Divide them by 2 and you will have: 10 pairs of X + Y
6 P 3 ==120 permutations. Same as above.
