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7 people, A, B, C, D, E, F and H, go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. In how many different arrangements will there be at least one person between A and F?

User Tuergeist
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1 Answer

4 votes

Answer:

There are 30240 ways the 7 people can be arranged

Explanation:

When calculating the "at least" value, this means we are looking for the total number of arrangements minus the number of arrangements where A and F are seated together.

To calculate the total number of ways, we use factorial (!)

Total arrangements = 8 x 7 x 6 x 5 x 4 x 3 x 2 = 40320

Together arrangements = 7 x 6 x 5 x 4 x 3 x 2 = 5040 (But A and F can swap seats, so the actual value is double what we calculated)...

5040 x 2 = 10080

Therefore the at least value can be calculated:

40320 - 10080 = 30240

User Adamprocter
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