arnaudl wrote:

Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?

A. 1/5

B. 1/4

C. 3/8

D. 2/5

E. 1/2

I think the best combinatorial solution was given by

heyholetsgo. Neat and short!

I didn't see a probabilistic approach, so I tried to work out one. Here it is:

There are 5 chairs, let's visualize them _ _ _ _ _

There is one single person, let's denote him/her by S.

S can choose from 5 chairs where to sit. There are some symmetrical placements for S:

1) S _ _ _ _ and _ _ _ _ S (0, 1, or 2 couples can be placed together)

2) _ S _ _ _ and _ _ _S_ (0 or 1 couple can be placed together)

And there is the third case:

3) _ _ S _ _ (0, 1, or 2 couples can be placed together).

So, let's compute for each case the probability that 0 couple sits together.

1) S _ _ _ _ or _ _ _ _ S:

\(\frac{2}{5}\) probability that S sits on either chair 1 or chair 5.

Now start placing people on the remaining four chairs and express the corresponding probabilities:

\(\frac{4}{4}\) for the first chair in the sequence of four remaining chairs

\(\frac{2}{3}\) for the second chair (cannot be the first person's mate)

\(\frac{1}{2}\) for the third chair (cannot be the mate of the second person)

\(\frac{1}{1}\) for the last person to be placed

In conclusion, the probability of no pairs sittings together when S sits on either chair 1 or 5 is given by

\(\frac{2}{5}*\frac{4}{4}*\frac{2}{3}*\frac{1}{2}*\frac{1}{1}=\frac{2}{15}.\)

2) _ S _ _ _ or _ _ _ S _

Again, two possibilities for S, so \(\frac{2}{5}\) probability for S to sit on either chair 2 or 4.

Now start placing people on the remaining four chairs and express the corresponding probabilities.

Treat only the _ S _ _ _ case, the other one has the same probabilities:

\(\frac{4}{4}\) for the first chair from the left

\(\frac{2}{3}\) for the second chair (cannot be the first person's mate because then the last two chairs will be occupied by the other couple)

\(\frac{1}{2}\) for the third chair (cannot be the mate of the second person)

\(\frac{1}{1}\) for the last person to be placed

In conclusion, the probability of no pairs sittings together when S sits on either chair 2 or 4 is given by

\(\frac{2}{5}*\frac{4}{4}*\frac{2}{3}*\frac{1}{2}*\frac{1}{1}=\frac{2}{15}.\)

3) _ _ S _ _

Here now we just have one possibility to place S, so \(\frac{1}{5}\) probability for S to sit on chair 3.

Starting from the left, place people on chairs:

\(\frac{4}{4}\) for the first chair from the left

\(\frac{2}{3}\) for the second chair (cannot be the first person's mate)

\(\frac{2}{2}\) for the third chair (we are already assured that we separated the two couples)

\(\frac{1}{1}\) for the last person to be placed

In conclusion, the probability of no pairs sittings together when S sits on chair 3 is given by

\(\frac{1}{5}*\frac{4}{4}*\frac{2}{3}*\frac{2}{2}*\frac{1}{1}=\frac{2}{15}.\)

In conclusion, the requested probability is \(3*\frac{2}{15}=\frac{2}{5}.\)

Answer D

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PhD in Applied Mathematics

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