In how many ways can we arrange 16 persons in 8 buildings and each building should have only 2 persons

Question

In how many ways can we arrange $16$ persons in $8$ buildings so that each building has only $2$ people? My answer is $16!/(2^8)$ but I'm not sure.

Answer 1

Your answer is correct:

Firstly, put all $16$ people in line like this:

$$A_1A_2A_3A_4\cdots A_{16}$$

The line can also be:

• First person in building $1$

• Second person in building $1$

• First person in building $2$

• Second person in building $2$

• $\cdots$

• First person in building $9$

• Second person in building $8$

Of course there are $16!$ ways of putting them in line.

Now, sort them into eight groups:

$$A_1A_2|A_3A_4|A_5A_6|A_7A_8|A_9A_{10}|A_{11}A_{12}|A_{13}A_{14}|A_{15}A_{16}$$

We will take just one case: $A_1$ and $A_2$ in building $1$, $A_3$ and $A_4$ in building $2$,... $A_{15}$ and $A_{16}$ in the building $8$.

• $A_1$ and $A_2$ in building $1$: two ways to switch them in building $1$.

• $A_3$ and $A_4$ in building $2$: two ways to switch them in building $2$.

$\cdots$

• $A_{15}$ and $A_{16}$ in the building $8$: two ways to switch them in building $8$.

So there are $2^8$ ways to swap pairs of people around just their buildings for each case, but we only count once for each case, so the answer is $16!/2^8$.