Dan, Ben, and Mark are friends. Together, along with two of their classmates, they bought movie tickets that were all in one row. How many different ways can everybody sit if Dan, Ben, and Mark want to sit together?

Answer 1

There are 5*4*3*2*1 = 5! = 120 ways total to have them seated. Call every person A, B, C, D, and E.

Set A, B, and C as Dan, Ben, and Mark, and the two other people as D and E. Make A, B, and C variable X.

The number of ways of X, D, and E sitting together now is 3*2*1 = 3! = 6.

But don't forget, there are 3*2*1 = 3! = 6 ways of making X:
1. ABC
2. ACB
3. CBA
4. CAB
5. BAC
6. BCA

So, the final answer is 6*6 = 6^2 = 36 different ways for everyone to sit if Dan, Ben, and Mark want to sit together.