Permutation/Combination Problem

There are 5 seats. _ _ _ _ _

Let the single person be X,

When X occupies 1st; X _ _ _ _ (4*2*1*1) = 8
When X occupies 2nd; _ X _ _ _ (4*X(1)*2*1*1) = 8
X takes 3rd ; _ _ X _ _ (4*2*X(1)*2*1) = 16
4th and 5th are repetitions of 1st and 2nd cases;

Together we have 8+8+16+8+8 = 48.

Total Cases = 5! = 120

Hence required probability = 48/120

plz explain in detail, answer is 2/5

This problem has been discussed in some really old post.

I am not sure if I would be able it explain it as well as the experts do.

Pls have a look https://www.beatthegmat.com/war-of-the-r ... 12457.html Could ask me if you have doubts though

The way I would solve this is using the glue method (from manhattan gmat).


Since the couple should not sit next to each other, I will make them sit next to each other and then count.

For problems in which items or people must be next to each other, pretend that the
items "stuck together" are actually one larger item.

Case when Both couples sit together:
2 couples and 1 loner. So A and B(couple), C and D(couple) and E are the people.
Glue AB and CD as single people. SO, i have AB, CD and E. 3 people in question. 3 seats in place.

Possibilities are = 3! * 2(AB can interchange) * 2(CD can interchange). = 24

Case when one couple sit together:

AB,C,D,E
Possibilities = 4! * 2(A and B interchanging) = 48
The possibility that CD sit together and AB don't, need not be calculated.

So totally = 48+24=72
Total possible permutations = 5! = 120

Probability = (120 - 72) / 120 = 48/120 = 2/5

Hope this helps !

I posted a solution here:

https://www.beatthegmat.com/couples-prob ... 82365.html