6 unique experienced persons(adsbygoogle = window.adsbygoogle || []).push({});

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.

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# Elementary combinatorics problem - why am I wrong?

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