[quote="jfranco23"]Of the 3 digit positive integers that have no digits equal to 0, how many have 2 digits that are equal to each other and the remaining digit different from the other 2?
A. 24
B. 36
C. 72
D. 144
E. 216[/quote][color=darkblue][/color]
IMO, using Box approach is simpler for computing permutations. In the question, we have 3 digits. Lets say 3 boxes. Since the question states that 2 digits are equal, for computation purpose lets consider the 2 equal digits as one box. This gives us 2 boxes. In the first box, one of the 9 numbers can be assigned. This leaves the next box that can be assigned with 8 numbers.
Hence 9*8=72
The 2 equal digits can be placed in two ways e.g 989, 899 (note that the third way of placing is fliping the first combination, 989, is redundant). Multiply the above with 2; hence 72*2=144.
Answer is D - IMO.
