We are to compute the probability that in 4 independent choices among 4 options (say labeled A, B, C, D, with equal likelihood), exactly one option appears twice, and two other distinct options appear once each, with the fourth position filled by a fourth distinct option — but wait: this would require 4 distinct options total, and one repeated. Since only 4 positions exist, and we want exactly one option repeated, and the other two being different, the only valid pattern is: one option appears twice, two others appear once, and the fourth option appears zero times — but that’s only 4 choices total. So the pattern is: one option repeated twice, and two other distinct options appearing once each — that uses up 2 + 1 + 1 = 4 choices, with one option appearing twice and two others once each, and one option not used at all. So the multiset of choices is of the form A, A, B, C, where A, B, C are distinct, and D is unused. - Veritas Home Health