Probability

[cypherskull]

2 couples and a 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?

a. 1/5
b. 1/4
c. 3/8
d. 2/5
e. 1/2

[Bill@VeritasPrep]

I think it's going to be easier to find the number of ways in which they do sit together, then subtract from 1.

Let's call one couple A1-A2, the other couple B1-B2, and the single person C.

Ways in which both couples sit together:

If we treat each couple as a unit, then we have 3 units to arrange. 3! = 6, but for each couple we have two ways to arrange: A1-A2 & A2-A1, B1-B2 and B2-B1. This means that for each of those 6 arrangements, we can flip the people in couple A or couple B and create a new arrangement. 3! * 2! (ways to arrange A) * 2! (ways to arrange B) = 24.

Ways in which one couple sits together:

Let's use couple A as the one sitting together. We know have 4 units to arrange: A, B1, B2, and C. 4! = 24, but again, we can flip the people in couple A to create a new arrangement. 4! * 2! = 48. HOWEVER, this includes all of the arrangements where B1 and B2 end up sitting together as well. We know that number is 24, so we must subtract it. 48-24 = 24.

We also must remember that couple B could be the couple sitting together. For each of the couple A arrangements, we could swap them with couple B, giving us 24 more arrangements, for a total of 48 arrangements where exactly one couple sits together.

This means that we have a total of 24+48 = 72 ways where either one or both couples sit together. How many total ways can we arrange 5 people? 5! = 120. 120 - 72 = 48 ways in which no couples sit together.

P(no couples together) = 48/120 = 2/5.

[Anurag@Gurome]

2 couples and a 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?

a. 1/5
b. 1/4
c. 3/8
d. 2/5
e. 1/2

First let us find the probability of at least one of the couple being sitting together.
One single person and two couples implies in all there are 5 people, so no. of ways of seating these 5 people = 5! = 120 ways
Let us consider one couple as 1 person, then no. of ways of seating them = 4! ways and 1 couple means 2 people, who can be seated in 2! ways. So, no. of ways of seating them = 4! * 2! = 48 ways
Similarly, the other couple can be seated in 48 ways.
Hence, total no. of ways in which the 2 couples sit together = 3! * 2! * 2! = 24 ways
Now, the no. of ways so that at least one of the couple sit together = 48 + 48 - 24 = 72
Probability so that at least one of the couple sit together = 72/120 = 3/5

Therefore, probability that neither of the couples sits together in adjacent chairs = 1 - 3/5 = [spoiler]2/5[/spoiler]

The correct answer is D.

[bubbliiiiiiii]

Hi Bill,

My approach towards the solution is same as your's. But I don't get why are you considering this part?

Ways in which one couple sits together:

Let's use couple A as the one sitting together. We know have 4 units to arrange: A, B1, B2, and C. 4! = 24, but again, we can flip the people in couple A to create a new arrangement. 4! * 2! = 48. HOWEVER, this includes all of the arrangements where B1 and B2 end up sitting together as well. We know that number is 24, so we must subtract it. 48-24 = 24.

We also must remember that couple B could be the couple sitting together. For each of the couple A arrangements, we could swap them with couple B, giving us 24 more arrangements, for a total of 48 arrangements where exactly one couple sits together.

I am OK until here,

Ways in which both couples sit together:

If we treat each couple as a unit, then we have 3 units to arrange. 3! = 6, but for each couple we have two ways to arrange: A1-A2 & A2-A1, B1-B2 and B2-B1. This means that for each of those 6 arrangements, we can flip the people in couple A or couple B and create a new arrangement. 3! * 2! (ways to arrange A) * 2! (ways to arrange B) = 24.

I am confused then! Could you please elaborate your explanation?