The Scooby Doo gang is going to the movies... again! That\'s right, they haven\'

Question

The Scooby Doo gang is going to the movies... again! That's right, they haven't defriended Fred in real life (yet).

Enter your answers for this problem as integers.

Suppose a row of seats at the movies is 10 seats, and that this row will be filled up completely by the 5 members of the Scooby Doo gang, and 5 strangers. How many arrangements of these 10 people are possible, such that the Scooby Doo gang can all sit adjacent to one another in this row?

Answer 1

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Now suppose Fred is being a huge jerk, so the Scooby Doo gang decides he has to sit on one of the ends of their group of 5 (not necessarily an end of the row). How many possible arrangements of the 10 people in the row are there now?

Answer 2

Sanity check: Should your answer for this part be greater than or less than the answer from the first part?

Explanation / Answer

First part: Since all 5 members of Scooby Doo gang sit together, let us take the gang as one person.

There are 1 + 5 = 6 persons and they can be arranged in 6! = 720 ways.

Further, the members of the gang can sit in 5! = 120 ways.

=> Total number of seatings = 720 * 120 = 86400.

Second part: Since Fred sits at one of the ends, the remaining four form a 'small' person. The gang itself is taken as a person and so the 7 persons can as earlier be seated in 6! = 720 ways.

Within the gang however, Fred has 2 choices and the remaining 4 people can sit in 4! = 24 ways. The gang therefore can sit in 2 * 24 = 48 ways.

=> Total number of seatings = 720 * 48 = 34560.

Sanity check: Naturally the second part should have a smaller number as the result as there are more seating restrictions.

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