Question

On a weekend, two parents take their children (A, B and C) to the movies. They sit in a row of five seats. Find the probability that

a) A and B sit next to each other.

b) Alice and C do not sit next to each other.

c) Alice and B sit next to each other while A and C do not sit next to each other.

Answer 1

a As A and B would sit next to eachother then take them as a single entity.Now we have 4 entities to arrange,which can be done in 4! =24 ways.

Also A and B can arrange between themselves in 2 ways.

Hence total possible ways=24×2=48

b.now if Alice and C sat next to each other then the no of ways =48 (as earlier)

Now total possible ways to arrange the 5 people without any constraints =5!=120

Then Alice and C would not sit next to each other in 120-48=72 ways.

c.if we want Alice and B to sit next to eachother then take them as a single entity.

And if we also want A and C to sit next to eachother we take them as a single entity.

So now we have 3 entities which can be arranged in 3!=6 ways.

And as Alice and B can arrange between them ,there is 2! Ways to do that.

Also as A and C can arrange between themselves ,there is 2 ways to do that.

So in 6×2×2=24 ways Alice and B can sit next to eachother and A and C can sit next to eachother.

Now in 48 ways only Alice and B sir next to eachother.

Then in 48-24=24 ways Alice and B sit next to eachother but A and C don't sit next to eachother.