<p>This is #15 on page 657 in the blue book.</p>
<p>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 experience plumber and 2 trainees, how many different such teams are possible?</p>
<p>The answer is 24.
A B C D are experienced plumbers
W X Y Z are trainees</p>
<p>A + WX
A + WY
A + WZ
A + XY
A + XZ
A + YZ</p>
<p>There are 6 combinations for plumber A
Since there are 4 plumbers to choose from, 6 x 4 = 24.</p>
<p>Thanks, this is the fastest method right?</p>
<p>I prefer to use it. If you're good with combinations, you can use that. but I prefer just writing it out</p>
<p>Think of it this way.</p>
<p>3 spots: _ _ _ (A team consists of 1 experience and 2 trainee) so...</p>
<hr>
<pre><code> E T T (There are 4 experienced and 4 trainee to choose from so...)
4 4 3 (Note: the last one is 3 since u already used 1 of 4 plumbers)
</code></pre>
<p>So do 4 x 4 x 3 = 48 possible combinations, but WAIT!</p>
<p>This is a combination problem, where order doesn't matter. Since Trainee 1 and Trainee 2 are still both Trainees if u switch their positions, then there are some repeated spots.</p>
<p>So divided 48 by 2 since the two positions are identical. You get 48/2= 24 possible combinations for a "team."</p>
<p>I prefer the formulaic method b/c it requires little "critical thinking" from your part; you just have to apply the formula. Whereas simple counting is sometimes pretty time-consuming and not applicable in every circumstance.</p>
<p>Answer for WartonWanabe's question: P4 x 4C2 = 24 :D</p>
<p>I always forget to divide ;]</p>
<p>That's one disadvantage of counting method :D</p>