Question

Find the number of ways of arranging `n` people in a straight line, if two particular people must always be separated..?

Answer 1

`"Ways" = 2x((n-2)!), "where " x=sum_(y=1)^(n-2)y`

Explanation:

Let's work a few examples with a growing value of `n` and see if we can find a pattern. Let's assign A and Z to the two people who must be separated.

When `n=3`, there are 2 ways we can sit the people - AXZ and ZXA. Ordinarily, we'd be able to sit 3 people in `3! = 6` ways.

When `n=4`, we can sit A and Z in seats 1, 3; 1, 4; and 2, 4 - which is 6 different ways. The other two people can sit in 2! = 2 ways for each of the 6 ways we can seat A and Z, which is `6xx2=12` ways. Ordinarily, we could seat four people `4! = 24` ways.

When `n=5`, we can seat A and Z in seats 1, 3; 1, 4; 1, 5; 2, 4; 2, 5; 3, 5 - which is 12 different ways. The remaining three people can sit in `3! = 6` different ways, giving `12xx6=72` ways. Ordinarily, we'd seat 5 people in `5! = 120` ways.

When `n=6`, we can seat A and Z in seats 1, 3; 1, 4; 1, 5; 1, 6; 2, 4; 2, 5; 2, 6; 3, 5; 3, 6; 4, 6 - which is 20 ways. The remaining four people can sit in `4! = 24` different ways, giving `20xx24 = 480` ways. Ordinarily, we'd seat 6 people in `6! = 720` ways.

What can we find from the above?

• We can express the ways to seat the people other than A and Z as `(n-2)!`

• We can express the ways to seat A and Z as `2x`. It's the `x` that is the thing we need to express that is the problem.

When:
- `n=3, x=1`
- `n=4, x=3`
- `n=5, x=6`
- `n=6, x=10`.

And so we can express `x` as the sum of the natural numbers from 1 to `(n-2)`:

`x=sum_(y=1)^(n-2)y`

And that is the expression we can use to answer our question:

`"Ways" = 2x((n-2)!), "where " x=sum_(y=1)^(n-2)y`

Let's give this a quick test. For `n=6`, we have:

`2x((6-2)!), x=sum_(y=1)^(n-2)y=1+2+3+4=10`

`2(10)(4!)=20xx24=480`

For fun, let's try `n=10`:

`2x((10-2)!), x=sum_(y=1)^(n-2)y=1+2+3+4+5+6+7+8=36`

`2(36)(8!)=72xx40320=2,903,040`