seating arrangments with n chairs but the first chair must be a girl

Question

Take the example of 2 chairs: there are two possible arrangments as there must be a girl in the first chair, gb or gg
The problem requires you to find the number of girls in all arrangments ie. for 2 chairs there are and for 3 chairs there are 8. I was wondering whether there was a way to generalise the result.

Answer 1

There seem to be a lot of implicit assumptions in your question. If I understand correctly, you’re dealing with children, and you’re assuming that each of them can be categorized as a boy or girl. Since I don't share this assumption, I’ll solve the problem for arrangements of $€1$ and $€2$ coins.

If I understand correctly, you don’t have a fixed number of either type, you only want a $€1$ coin in the first slot, and all remaining slots can be filled with either coin. Again if I understand correctly, you seem to be looking for the total number of $€1$ coins summed over all possible arrangements.

If there are $n$ slots, since one of them is fixed, there are $2^{n-1}$ ways to fill the remaining ones. The first slot always contains a $€1$ coin, whereas the remaining $n-1$ slots do so exactly half the time. Thus, the total number of $€1$ coins in all arrangements is

$$ 2^{n-1}\left(1+\frac{n-1}2\right)=(n+1)2^{n-2}\;. $$

For $n=3$, this is indeed $(3+1)2^{3-2}=8$.

Answer 2

In all arrangements first place is occupied by girl and for rest of the places there would be girl or boy.$$ $$ If we consider total at n places let other than 1st place from rest (n-1) places r places are occupied by girl. $$ $$ Hence number of girl in that arrangement is $(r+1){{n-1} \choose r}$ $$ $$ Hence total number of girls are $$\sum_{r=0}^{n-1} (r+1){{n-1} \choose r}\,=\,(n+1)2^{n-2}$$