If 2 people out of 7 refuse to sit next to each other in a row of 7 seats, how many ways are possible?

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If 2 people out of 7 refuse to sit next to each other in a row of 7 seats, how many ways are possible?

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Answer 1 · 2018-02-24T08:06:48

$7! - 2 \cdot 6! = 3600$ $7! - 2 * 6! = 3600$

Explanation:

The total number of ways in which $7$ $7$ people can be seated in a row of $7$ $7$ seats is:

$7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 7! = 5040$ $7 * 6 * 5 * 4 * 3 * 2 * 1 = 7! = 5040$

To find the number of these ways that are unacceptable because they result in the particular pair being sat next to one another, we can think of seating $6$ $6$ people in a row of $6$ $6$ seats, excluding one person from the particular pair, then suppose they arrive and squeeze an extra seat into the row one side or the other of the other member of the pair.

That can be done in $2 \cdot 6! = 1440$ $2 * 6! = 1440$ ways.

So the total number of acceptable arrangements is:

$7! - 2 \cdot 6! = 5040 - 1440 = 3600$ $7! - 2 * 6! = 5040 - 1440 = 3600$

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If 2 people out of 7 refuse to sit next to each other in a row of 7 seats, how many ways are possible?

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Explanation:

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