<p>A plumbing company wants to send a team of 3 plumbers.
It has 4 experienced plumbers and 4 trainees available at its disposal.
If a team consists of 1 experienced plumber and 2 trainees, how many different teams are possible?</p>
<p>I said that there are 4X3 ways to choose the trainees and 4 ways to choose the experienced guy</p>
<p>for 48 different teams</p>
<p>but the answer key says 24...</p>
<p>C(4,2) = 6 (possible teams of 2 trainees)
6 * 4(experienced plumbers to choose from) => 24</p>
<p>Order does not matter for these teams.</p>
<p>^, i’m not even going to lie, I just learned combinations and permutations like last week in Alg II… ROFL. I’m glad my test isn’t for another 12 days, but yeah Violin’s got it I believe.</p>
<p>wait, can someone do this again. i keep getting confused between permutations and combination. this is a combination question correct? because order does matter. so, how do i apply the formula?</p>
<p>4 Plumbers, 4 Trainees. You want 1 Plumber, you have 4 choices.
4 x …</p>
<p>Next, you want 2 Trainees. (1,2,3,4)
This is a combination, because it does not matter if you choose trainee 1 before trainee three in the pair (1,3).</p>
<p>If you choose the 1st trainee, you could have (1,2) (1,3) (1,4)
If you choose the 2nd trainee, you could have (2,3) (2,4)
If you choose the 3rd trainee, you could have (3,4)</p>
<p>The formula is: (n)! / [ (n-r)! r!] where n is number you choose from, r is how many you are choosing. (n=4, r=2)
4! / [2! * 2!] = 6 </p>
<p>4*6=24.</p>
<p>Permutations are used when order does matter, and repetition is allowed (ie. List of possible 1st Place, 2nd Place, and 3rd Place winners in a race). Combinations are used when order does not matter and repetition is not allowed (ie. List of possible 3-member teams that are selected from a group of 10 people)</p>
<p>yup, thanks. The value for a question in which you can use permutation or combination (for the sake of my statement) will always yield a greater value for permutations, which logically makes sense.</p>