6 unique experienced persons 2 unique inexperienced persons How many 3-person combinations if at most one in the group can be inexperienced? 6*5*4/3! + 6*5*2/2! is the answer. My answer was 6*5*4/3! + 6*5*2/3! Why does the book divide by a 2-permutation and not a 3-permutation for the second term (all combinations consisting of 2 experienced persons and 1 inexperienced person)? My (flawed) rationale is that you have three slots to be filled; we have 6 experienced choices for the first slot, 5 remaining experienced choices for the second slot, and 2 inexperienced choices for the last slot. Since, we don't consider order, we divide by a correction factor of 3 slots permuted.