Problem Solving

pnk wrote:Hi Rahul,

We can also solve this by finding couples always sitting together and thn subtracting from 1.

Using this, could you find error in my approach. I am not able to find one:

5 people = individual 1 (I1) + couple 1 (C1)+couple 2 (C2) => 3 people only

I will try to find possibilities where c1 and c2 sit together. In that case, we have 3 persons sitting =3!. Plus each set of couple can arrange among themselves in 2! ways. Therefore nos of ways in which 5 people can sit (with couples sitting adjacent) = 3! 2! 2! = 24

Probability of not sitting together = (1 - 24/5!) = 4/5

Can someone help me where I am wrong

chaitanyareddy wrote:Hi Experts,

I am facing real tough time with permutation and combinations and also probability. anyone of you , if you are having a good ebook or notes regarding P and C and probaility , please post the same..

Could you please explain me the solution for this problem.

Q. TWO couples and a single person are to be seated on 5 chairs such that no
couple is seated next to each other. What is the probability of the above??

GMATGuruNY wrote:
We could treat this as an overlapping groups problem:

Total = G(1) + G(2) - Both + Neither

Let's say we have couple AB, couple CD, and lonely person E.

Total arrangements = (arrangements with AB together) + (arrangements with CD together) - (arrangements with both AB and CD together) + (arrangements with neither AB nor CD together)

Total arrangements = 5! = 120
Total arrangements with AB together means we're arranging the 4 elements AB, C, D, and E = 4!*2 = 48 (We multiply by 2 since AB can be reversed to be BA.)
Total arrangements with CD together means we're arranging the 4 elements CD, A, B and E = 4!*2 = 48 (We multiply by 2 since CD can be reversed to be DC.)
Total arrangements with AB and CD together means we're arranging the 3 elements AB, CD, and E = 3!(4) = 24 (We multiply by 4 because AB can be reversed, CD can be reversed, and both AB and CD can be reversed.)

Plugging into the overlapping groups formula, we get:

120 = 48 + 48 - 24 + N
120 = 72 + N
N = 48

So P(neither couple sitting together) = 48/120 = 2/5.

Probability of couple never being together = 1 - Probability of couple always together.

Couple AB, CD; Individual E
---->U have forgotten something here. This as well includes AB,C,E,D means only one couple is treated as one unit and other not...But this scenario also shld be excluded as the problem says no couple shld sit together...

Arrangement with AB, CD together = 3! 2! 2! (3! bcoz AB, CD, E is 1 group each - and these 3 groups can be arranged among themselves in 3! ways; 2! each for AB and CD arranging among themelves) = 24

So, probability = 1 - 24/120 = 4/5. Could you pls explain where I am at fault.

pnk wrote:Hi Mitch,

I agree with your logic but why my approach gives me a different answer:

Probability of couple never being together = 1 - Probability of couple always together.

Couple AB, CD; Individual E

Arrangement with AB, CD together = 3! 2! 2! (3! bcoz AB, CD, E is 1 group each - and these 3 groups can be arranged among themselves in 3! ways; 2! each for AB and CD arranging among themelves) = 24

So, probability = 1 - 24/120 = 4/5. Could you pls explain where I am at fault.

GMATGuruNY wrote:

pnk wrote:Hi Mitch,

I agree with your logic but why my approach gives me a different answer:

Probability of couple never being together = 1 - Probability of couple always together.

Couple AB, CD; Individual E

Arrangement with AB, CD together = 3! 2! 2! (3! bcoz AB, CD, E is 1 group each - and these 3 groups can be arranged among themselves in 3! ways; 2! each for AB and CD arranging among themelves) = 24

So, probability = 1 - 24/120 = 4/5. Could you pls explain where I am at fault.

You need to account for all the different ways that one or both couples could sit together. Here's an equation that you could use:

P(neither couple together) = 1 - P(only AB together) - P(only CD together) - P(both AB and CD together)

Number of ways to arrange AB, CD and E = 3! * 4 = 24.
So P(AB and CD together) = 24/120 = 1/5.

Number of ways to arrange AB, C, D and E with CD not together = (Total arrangements of AB, C, D, and E) - (number of arrangements with both AB and CD together) = (4! * 2) - 24 = 24.
So P(only AB together) = 24/120 = 1/5.

Number of ways to arrange CD, A, B and E with AB not together = (Total arrangements of CD, A, B, and E) - (number of arrangements with both AB and CD together) = (4! * 2) - 24 = 24
So P(only CD together) = 24/120 = 1/5.

Thus P(neither couple together) = 1 - 1/5 - 1/5 - 1/5 = 2/5.

Hope this helps!