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$10$ Boys and $10$ Girls get ordered in a line. How is $X$, the number of boys standing between the leftmost girl and the rightmost girl, distributed?

I tried thinking of selecting one place from the $20$ for the leftmost girl, and then selecting k places from the $19$ left for the boys. Or selecting one place from the $19$ left for the rightmost girl. I can't figure how to solve this.

Any help is appreciated.

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  • $\begingroup$ Consider the possible number of positions between the leftmost girl and rightmost girl. $\endgroup$ Jun 26 at 9:53

2 Answers 2

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The total number of placements is $\binom{20}{10}$. The number of placements with $n$ boys between the outermost girls is $(11-n)\binom{8+n}8$, namely $11-n$ ways for the pair of outermost girls to be placed, and $\binom{8+n}8$ ways to place the remaining $8$ girls in the places in between. So the distribution is given by those numbers for $n=0,\ldots,10$ over a common denominator of $\binom{20}{10}=184756$. The respective numerators are $$11,90,405,1320,3465,7722,15015,25740,38610,48620,43758$$ and indeed their sum is $184756$.

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  • $\begingroup$ Why is it okay to think only about positioning the girls? Is it because for the boys we pick n places from n? $\endgroup$ Jun 27 at 6:38
  • $\begingroup$ @topzeramail Once you know which positions are occupied by girls, those at the remaining positions must be boys; there is no choice (or more precisely, just one possible choice) left. I guess this glosses over the possibility of non-binary kids, but that omission is more or less enshrined in the question. $\endgroup$ Jun 27 at 10:20
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    $\begingroup$ What is a "non-binary kid"? $\endgroup$ Jun 27 at 11:44
  • $\begingroup$ @John_Krampf A kid that does not identify itself with either the (pure) female or male gender $\endgroup$ Jun 27 at 11:53
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    $\begingroup$ Is this an American thing? I never heard of them before $\endgroup$ Jun 27 at 14:04
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Between the leftmost girl and the rightmost girl are always 8 girls plus $0$ to $10$ boys.

Now:

  • If there is no boy between them, the group of $\color{red}{10}$ girls may be in 11 different configurations:

    [GGGGGGGGGG]bbbbbbbbbb
    b[GGGGGGGGGG]bbbbbbbbb
         ...
    bbbbbbbbbb[GGGGGGGGGG]
    
  • If there is 1 boy, the group from the leftmost girl to the rightmost girl will have $\color{red}{11}$ members, and there are only 10 different places for the group as a whole, for example

    bb[GGGGGGbGGGG]bbbbbbb
    

    On the other hand, every of 10 possible locations may have that 1 boy in 9 different positions inside a group, so there are 10 × 9 = 90 possibilities.

  • If there are 2 boys, the group from the leftmost girl to the rightmost girl will have $\color{red}{12}$ members, and there are only 9 different places for the group as a whole.

    But every of 9 possible locations may have those 2 boys in $\binom{10} 2$ = 45 different positions inside a group, so there are 9 × 45 = 405 possibilities.

So the distributions of boys between the leftmost and the rightmost girl is for now:

Boys Occurrences                                                                                                   
0 11
1 90
2 405

And you may continue ...

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  • $\begingroup$ The boys cannot be at an outermost position in the group. $\endgroup$ Jun 26 at 12:10
  • $\begingroup$ @MarcvanLeeuwen, thank you very much, you are right, I corrected my answer. $\endgroup$
    – MarianD
    Jun 26 at 12:15

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