<p>Q: 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 consist of 1 experienced plumber and 2 trainees, how many different such teams are possible? </p>
<p>ANSWER: 24</p>
<p>Okay. So i did 4 x 4 x 3 because a team consist of one experienced plumber so that is 4, and 2 trainees, 4 and 3.. </p>
<p>I thought that is how we solve these types of questions? Please help!! And Thanks </p>
<p>Think of it this way: there are four ways to pick the 1 experienced plumber (4 experienced plumbers, pick 1 at a time) and 6 ways to pick the two trainees (4 trainees, 2 at a time). If you label the trainees A B C D, you have AB, AC, AD, BC, BD, and CD. So, 4 x 6 = 24. </p>
<p>@Parampreet You have to divide by 2, because picking trainees A and B is the same as picking B and A.</p>
<p>In more complete terms, the number of ways is (4C1)(4C2) = 4*6 = 24, where nCk (\binom{n}{k}) = n!/((n-k)!k!) is the number of ways to choose a subset of k objects from n with ordering irrelevant.</p>