In how many ways can you put two people in $5$ seats in a row if they cannot sit together?

Question

In how many ways can you put two people in $5$ seats in a row if they cannot sit together?

So this is a simple combinatorics problem that i can't quite work out. I know that the answer is $12$ but i don't fully understand how to get it.

Answer 1

There are $5 \cdot 4 = 20$ ways to sit them in $5$ seats without any restrictions.

And there are $4\cdot 2 = 8$ ways to sit them together ($4$ places to sit together, $2$ options for switching two people).

So the answer is $20-8 = 12$.

Answer 2

Hint:

$$\text{Find the number of ways they can sit (without restrictions).}\tag1$$

$$\text{Then, find the number of ways they cannot sit.}\tag2$$

$$\text{Answer}=(2)-(1)$$

Answer 3

Another way similar to ArsenBerk's method is grouping the two people together to become one. There are $4*3$ ways to do this, and you just subtract that from the total ways which is $5*4$.

Answer 4

If we number the seats from $S_1$ to $S_5$, here are the possibilities:

$S_1$ for the first person. For the second person, there are $3$ possibilities.

$S_2$ and $2$ possibilities.

$S_3$ , with $2$

$S_4$ with $2$ finally $S5$ and $3$

$$\text{total}=3+2+2+2+3=12$$

Answer 5

The number of ways to seat $2$ people in $5$ seats is the permutation $5$p$2$ since order matters. $\frac{120}6 = 20$. Now, how many ways can they sit together? To figure this out I put $5$ dots on my paper and circled every possible pair. You can only do that $4$ times. There may be a more mathematical way to do that though. There are two ways to get a pair, which is $2!$. So, $4\cdot 2! = 8$ possible ways to sit together.

Therefore, $20 - 8 = 12$.