<p>Test 7 Section 3 Number 15.
The Acme Plumbing Company will send a team of 3 plumbers to work on a certain job. The company has 4 experienced plumbers and 4 trainees. If a team consists of 1 experienced plumber and 2 trainees, how many different such teams are possible?
We know how to solve by showing that each experienced plumber has a 6 possible combinations, multiplied by 4 experienced plumbers- gives 24 different team possibilities.
My daughter tried the factorial method- 4 possible experienced (to choose from) x 4 possible (to choose from) trainees, x 3 trainees to fill in the third team member slot = 48 which is too many and incorrect.
I do not see how & why to tell her this method does not work. Thank you in advance for any light you can shed on this problem.</p>
<p>Order doesn’t matter for trainees. Trainee 1 and trainee 2 is the same as trainee 2 and trainee 1. You end up double counting the number of real combinations, have to divide by 2.</p>
<p>You are choosing 2 trainees from a set of 4, so the number of ways to do it (order doesn’t matter) is (4*3)/2! = 6. In general, the number of ways to choose k objects from a set of n, where order is irrelevant, is n!/((n-k)!k!).</p>