Problem Solving

5 people, 2 couples and one single are to be seated in such a way tht couples don't sit next to each other. Wht is the probability of tht happening.

IMO 5*3*2*2*1 = 60 if they are in a straight line or in a straight row.

Probability???

Probability that the couples are not sitting next to each other = 1 - Probability that the couples are sitting next to each other

Probability that the couples are sitting next to each other
Lets assume one couple is B1&B2 and the second couple is C1&C2 & the single is denoted by S. The number of ways we can seat them is shown below:

S C1 C2 B1 B2
C1 C2 S B1 B2
C1 C2 B1 B2 S

3 ways as shown above. Now, C1 & C2 can be arranged in 2 ways. Similarly B1 & B2 can be arranged in 2 ways. So, totally, the number of ways that the couples can sit together is = 2*2*3* = 12 ways

The number of ways both the couples & the single can be seated is = 5!/(2!*2!) = 30 ways.

So, the probability is 12/30 ==> 2/5ways

Therefore the probability of the couples not sitting together is 1-(2/5) = 3/5ways

Vemuri wrote: 3 ways as shown above. Now, C1 & C2 can be arranged in 2 ways. Similarly B1 & B2 can be arranged in 2 ways. So, totally, the number of ways that the couples can sit together is = 2*2*3* = 12 ways

The number of ways both the couples & the single can be seated is = 5!/(2!*2!) = 30 ways.

So, the probability is 12/30 ==> 2/5ways

Therefore the probability of the couples not sitting together is 1-(2/5) = 3/5ways

can you explain a little bit about the 30 ways?

thanks