**5(b)**

We can set the Earthlings in $n!$ ways if they are in straight line. But considering rotation, there exists a group of $n$ arrangements, that are actually same if we rotate them.

If $n=3$, rearrangements are: $ABC,BCA,CAB,ACB,CBA,BAC$, where the first three are same and so are the last three.

So, actual arrangements=$2$=$\frac{3!}{3}$. We can say:$\frac{n!}{n}$

The rest is same. $m$ Martians can be placed in $n$ places in $n \choose m$ ways. And, the Martians can be arranged in $m!$ ways.

So, the result=$\frac{n!}{n}.m!.{n \choose m}$=$\frac{n.{(n-1)!}^2}{(n-m)!}$

If $n=3$, rearrangements are: $ABC,BCA,CAB,ACB,CBA,BAC$, where the first three are same and so are the last three.

So, actual arrangements=$2$=$\frac{3!}{3}$. We can say:$\frac{n!}{n}$

**[Can check for $n=4$]**The rest is same. $m$ Martians can be placed in $n$ places in $n \choose m$ ways. And, the Martians can be arranged in $m!$ ways.

So, the result=$\frac{n!}{n}.m!.{n \choose m}$=$\frac{n.{(n-1)!}^2}{(n-m)!}$