Combinatorics: In how many of the ways can 2 specific people out of 3 people be seated, without an empty seat between them?

(Check out my solutions to these math problems below the questions)

Kimberly, Jack and Sofie are waiting for the bus.

1. In how many ways can these 3 create a line together?

When the bus arrives there are 5 seats next to each other at the end of the bus.

1. In how many different ways can Kimberly, Jack and Sofie sit on these 5 seats?

2. And in how many of those ways (I guess that it’s the ways from question 2) can Kimberly and Jack sit next to each other without an empty seat between them?

Solution for question 1: $$3! = 6$$ ways to create a line

Solution for question 2: $$\binom{5}{2} = 10$$ different ways to pick 3 out of 5 seats to sit on.

By using the multiplication principle we get that $$10 \cdot 6 = 60$$. So in total there are $$60$$ different ways for these 3 people to sit.

Solution for question 3: I could need help with this one, I find it tricky. If I do it graphically, counting all the times Kimberly and Jack sit together out of those $$60$$ ways from solution 2, I get it to be $$24$$. How would you answer this question? Perhaps there is an easier and better way.

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The block of two seats in which Kimberly and Jack sit must begin in one of the first four places. Choose whether Kimberly or Jack sits in the leftmost seat within the block. Choose in which of the remaining three seats Sofie sits. There are $$4 \cdot 2 \cdot 3 = 24$$ admissible seating arrangements.