In how many ways can 7 people be seated in a row o chairs if Jane and Joe must sit next to each other?

Jan 21, 2018

Seven people can be seated in 6!*2! =1440 ways, if two of them must be seated together.

Explanation:

This is a question about permutations. A permutation is an arrangement of a certain amount of objects in a specific order.

For example, if we have two people, $A$ and $B$, we can arrange them in a row of two chairs in two ways

$A B$ or $B A$

Each of these two arrangements is a permutation. Thus, there are only two possible permutations in the example.

If we have seven people it gets a bit trickier. We must use the multiplication principle.

If we have seven people and we want to know how many ways we can arrange them in a row of seven chairs, using the multiplication principle, we multiply all the options together to get the total number of arrangements. E.g. there are 7 options for the first seat, 6 for the second (because one has been used), 5 for the third, 4 for the fourth, and so on.

7*6*5*4*3*2*1=7! =5040

So there are 5040 ways of arranging seven people in a row of seven chairs.

I think the proof of this is best seen by making a tree diagram and noting that the number of branches at the end tells you the number of possible arrangements/permutations. The tree for seven people is way too big so if you want to test this, do it with a smaller number, such as three people (there will be $3 \cdot 2 \cdot 1 = 6$ branches/arrangements). I won't go into more detail on how this works, other than to say that it is very important and should be understood if you want to deal with permutations or probability.

Now, let's call each person $A , B , C , D , E , F , G$

The answer is not 5040 because we have a restriction, two of the people must sit next to each other. We'll call Jane and Joe, $A$ and $B$. Group them together and call them one object, $X$.

Now we effectively have six objects

$X , C , D , E , F , G$

where $X = A , B$

Using the multiplication principle, we can arrange six objects in 6! ways

6*5*4*3*2*1=6! =720

But we also have to take into consideration the group, $X$, which can also be arranged in multiple ways

$A B$ or $B A$

2*1=2! =2

To take this into account and get the total number of arrangements, we must use the multiplication principle again. This involves multiplying the number of arrangements of six objects by the number of arrangements of the group, $X$

$720 \cdot 2 = 1440$

This makes sense because we have 720 arrangements where $A$ is first, but we can have another 720 arrangements where $B$ is first.