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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.

arvass
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  • Why 8 for three chairs? – almagest Jan 07 '20 at 14:17
  • It sounds as though you have an unlimited supply of identical girls and an unlimited supply of identical boys. That goes against the common theme of problems like this where we usually have a specific number of available girls and a specific number of available boys, often the same number of people total as we have chairs, and where each person is unique. – JMoravitz Jan 07 '20 at 14:18

2 Answers2

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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 $(3+1)2^{3-2}=8$.

joriki
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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 $$\sum_{r=0}^{n-1} (r+1){{n-1} \choose r}\,=\,(n+1)2^{n-2}$$

mathsdiscussion.com
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