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Probability

Expert replies
by abhirup1711 » Tue Jul 09, 2013 4:42 am
Two couples and one single person are seated at random in a row of 6 chairs. What is the probability that neither of the couples sits together in adjacent chairs?
1/5
1/4
3/8
2/5
½
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Source: — Data Sufficiency |

by Matt@VeritasPrep » Tue Jul 09, 2013 4:25 pm
abhirup1711 wrote:Two couples and one single person are seated at random in a row of 6 chairs. What is the probability that neither of the couples sits together in adjacent chairs?
1/5
1/4
3/8
2/5
½
Let's say that our couples are (A,B) and (C,D), our single person is E and our empty chair is X.

First, we'll find the total number of arrangements: 6 "people" (we'll say the empty chair is a person) in 6 seats is just 6!, or 720.

Now let's find the odds that A and B sit together. To do this, let's treat them as one person. That means we have 5 elements to arrange, which gives us 5!. But A and B could be arranged "together" in 2 ways, so we have a total of 2 * 5!, or 240 ways in which they sit together.

We also have 240 ways in which C and D (the other couple) sit together.

However, if we do 240 + 240, we've DOUBLE COUNTED the arrangements in which both A and B and C and D sit together, as they appear in the A + B arrangements AND the C + D arrangements. Since we've double counted these arrangements, we have to subtract them once from our 240 + 240 total.

If both A and B and C and D sit together, we have 4! * 2 * 2 arrangements, or 96 arrangements with each couple together.

So, in total, we have 240 + 240 - 96, or 384 arrangements in which at least one couple is sitting together. That means we have 720 - 384, or 336 arrangements in which neither couple is sitting together.

That means the odds of neither couple together are 336/720, or 7/15.

However, since this isn't an answer, I'm guessing that your question may have meant to say "Two couples and one single person are seated at random in a row of FIVE chairs". In this case (using the logic we just used above, with slightly different numbers), the answer is 2/5, which is one of the answer choices.

(Sketch of the solution: Total = 5!. A + B sit together in 4! * 2 ways, C + D sit together in 4! * 2 ways, A + B and C + D sit together in 3! * 2 * 2 ways, so there are a total of 4! * 2 + 4! * 2 - 3! * 2 * 2 ways in which at least one couple is sitting together, or 72/120. But we want neither together, so 48/120 = 2/5.)
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