<p>Page 657</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 experienced plumber and 2 trainees, how many different such teams are possible?</p>
<p>HighlandDaughter came up with 4x4x3=48</p>
<p>Correct answer (according to BB) is 24.</p>
<p>??? Help please - thanks!</p>
<p>4x4x3 is ALMOST right!</p>
<p>But alas...</p>
<p>Say the plumbers are named A,B,C and D.
Say the apprentices are named w,x,y,and z.</p>
<p>You could start by saying you have 4 choices for the plumber, 4 for the first apprentice and 3 for the next. That's probably how she got 4x4x3.</p>
<p>But that method ends up counting Axy and Ayx as two different teams. It counts every team twice. So you have to divide your answer in half.</p>
<p>Or...</p>
<p>You could list the possible teams.<br>
Starting with plumber A, the teams are: Awx, Awy, Awz, Axy, Axz and Ayz. That's 6. But there will be 6 each for B, C and D as well.</p>
<p>think of it as combinations. The high-level combinatinos/permulation was always involve two different conditions/sets that need to be taken care of. In this case it's a 3 man team of plumber and trainees. Take the combination of each individual event and multiply.
4C2 * 4C1</p>
<p>Out of 4 trainees you want to see the number of possible sets possible of 2 where order doesn't matter
Out of 4 plumbers you want to see the number of possible sets possible of 1 experienced plumber where order doesn't matter.</p>
<p>After that you treat it like most permutations and combinations and multiply the two individual events together for one result.</p>
<p>Yeah, this was rather easy if you used your calculator. All I did was plug in the combination of the first one (n=4, r=2) and then multiply it by the second one (n=4, r=2).</p>
<p>4 apprentices, choose 1 => 4 choices
4 trainees, choose 2 => 4!/(2!<em>2!) = 6 choices
4</em>6=24.</p>
<p>It makes sense to check the <a href="http://talk.collegeconfidential.com/sat-preparation/339734-consolidated-list-blue-book-math-solutions-3rd-ed.html%5B/url%5D">http://talk.collegeconfidential.com/sat-preparation/339734-consolidated-list-blue-book-math-solutions-3rd-ed.html</a> whether the question in question has been already answered.
[ul]657 / 15 /[/ul] - check!</p>
<p>But there is something odd about this problem that makes me wonder if it was ever on an actual SAT. (I don't think all of the blue book tests were really used as tests -- not sure if any of them were.)</p>
<p>Here's what I mean: a confused student could say "Well the biggest number of choices is 4. And these kinds of problems usually involve factorials. So I'll try 4! = 24 as my answer."</p>
<p>You are not supposed to be able to get a hard problem right for the wrong reason. And they could have easily avoided this by changing the question so that there were 5 plumbers instead of 4. That would not make it harder but you wouldn't be able to stumble onto the right answer.</p>
<p>Thanks everybody. Very helpful. And gcf, I tried to search for the page number but it was too small (fewer than 10 characters).</p>