prob good one

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? 

Prob. that neither couples sit together = 1 - prob. that either of the couples sit together. 

Let AB - first couple 
CD - second couple 
E - single fellow. 

P(A or B) = P(A)+P(B)-P(A and B) 

P(A) = prob. of first couple sitting together. 
P(B) = prob. of 2nd couple sitting together. 
P(A and B) = prob. of both couples sitting together 

Take AB as one unit. So the number of ways of arranging 4 units = 4!. 
AB can be arranged in 2! ways. 
Hence number of ways first couple can sit together = 2*4! = 48 

Same for couple CD. 
Number of ways 2nd couple can sit together = 2*4! = 48 

Number of ways both couples can sit together. 
Same logic as above. Consider couple AB and couple CD as 2 units. 
So number of ways of arranging 3 units = 3! 
AB and CD can be arranged in 2*2 ways (AB in 2 ways and CD in 2 ways) 
both couples can sit together in 2*2*3! = 24 ways 

Put it into the formula. 
(A or B) = A + B - (A and B) 
48+48-24 = 72 

Total number of ways of arranging 5 people = 5! = 120 

So prob. that either couples sit together = 72/120 = 3/5 
Prob. that neither couples sit together = 1-3/5 = 2/5