<p>There are twenty couples at a high school dance. Each person has to get into groups of three. People who are a couple cannot be in the same group together. So how many 3 member groups can there be where no one in the 3 member group is a couple?</p>
<p>I tried ordering it like couple Aa, couple Bb, Couple Cc, etc. Then I put them in groups of three by doing ABC, Abc, ABc, etc. Can anyone help me here. I really think that I am kind of doing this problem right but is there any shortcut to this problem. I have written down so many combinations.</p>
<p>Think of it as a permutation problem at first (for simplicity). You then need to make choices for slot1, slot2 and slot3. For slot1, you have 40 choices; for slot2, you have 39-1=38 choices (since the significant<em>other of slot1 cannot be considered); and for slot3, you have 38-2 choices (since the significant</em>others of slot1 and slot2 cannot be considered).</p>
<p>So, if this was a permutation problem, the answer would be (40)(38)(36). However, since the order of the slots doesn't matter, the actual answer is (40)(38)(36) / (3)(2) , since each combination of ABC would yield 6 permutations ABC, ACB, BAC, BCA, CAB, CBA. </p>
<p>Final answer is (40)(38)(6). Good thing you quit writing down all possible 3_groups, you've saved a forest :) .</p>