Problem Solving

one single person and two couple are to be seated at random in a of raw five chairs ,what is the probability that neither of the couples sits together in adjacent chairs .

Md.Nazrul Islam wrote:one single person and two couple are to be seated at random in a of raw five chairs ,what is the probability that neither of the couples sits together in adjacent chairs .

Let's say that 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)

The big idea is to SUBTRACT THE OVERLAP.
When we count the arrangements in which AB sit together and those in which CD sit together, the arrangements in which both AB and CD sit together -- the OVERLAP -- gets counted twice.
Thus, we need to SUBTRACT THE OVERLAP -- the arrangements in which both AB and CD sit together -- so that these arrangements are not double-counted.

Total arrangements = 5! = 120

AB together:
Number of ways to arrange the 4 elements AB, C, D, and E = 4! = 24.
Since AB can be reversed to BA, we multiply by 2:
2*24 = 48.

CD together:
Number of ways to arrange the 4 elements CD, A, B and E = 4! = 24.
Since CD can be reversed to DC, we multiply by 2:
2*24 = 48.

Both AB and CD together:
Number of ways to arrange the 3 elements AB, CD, and E = 3! = 6.
Since AB can be reversed, CD can be reversed, and both AB and CD can be reversed, we multiply by 4:
4*6 = 24.

Plugging these values into the equation above, we get:

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

Thus, P(neither couple sit together) = 48/120 = 2/5.

Hi guys, first post for me on the boards so excuse any missteps as I make them. First, thanks a lot for working these solutions. I'd like to work through a problem using different numbers in order to solidify the concept. These numbers will never be used on the GMAT, but again I'm looking to practice the conceptual side of things with some guidance. Thanks!

Concept Problem: Now, let's say 4 more guests show up, another couple and two loners. So total there are 9 chairs for 9 guests - 3 couples and 3 stags. If the guests are seated randomly, what is the new probability that none of the couples will sit together?

Guest list: AB, CD, EF, G, H, I

Total arrangements = (arr. AB together) + (arr. CD together) + (arr. EF together) - (arr. AB + CD together) - (arr. AB + EF together) - (arr. CD + EF together) - (arr. AB + CD + EF together) + (arr. none are together) = 9! = 362, 880

AB together: We have 8 elements - AB, C, D, E, F, G, H, I - so total arrangements = 8! = 40, 320. Multiply by 2 to account for AB->BA switch = 80,640.
CD together: same reasoning = 80,640.
EF together: same reasoning = 80,640.

ABCD together: We have 6 elements - ABCD, E, F, G, H, I - so total arrangements = 6! = 720. Multiply by 2 to account for AB->BA switch, and by another 2 for the CD->DC switch = 720*4 = 2,880.
ABEF together: same reasoning = 2,880.
CDEF together: same reasoning = 2,880.

ABCDEF together: We have 4 elements - ABCDEF, G, H, I - so total arrangements = 4! = 24. Multiply by 2 for each couple's switch = 24 * 8 = 192.

Return to original: Total arrangements = (arr. AB together) + (arr. CD together) + (arr. EF together) - (arr. AB + CD together) - (arr. AB + EF together) - (arr. CD + EF together) - (arr. AB + CD + EF together) + (arr. none are together) = 80,640 + 80,640 + 80,640 - 2,880 - 2,880 - 2,880 - 192 + (arr. none are together) = 362, 880

Arrangements none together = 129,792

Probability none sit together = 129,792/362,880 = .357

Don't worry about running the actual numbers, I just want to check the conceptual framework. Thanks!