Row of Boys and Girls!

Probability Level pending

Suppose that 7 boys and 13 girls line up in a row. Let S S be the number of places in the row where a boy and a girl are standing next to each other. For example, for the row G B B G G G B G B G G G B G B G G B G G GBBGGGBGBGGGBGBGGBGG , we have S = 12 S=12 . The average value of S S (if all possible orders of these 20 numbers are considered) is close to which of the following natural numbers?

12 10 11 9

Walid Mustafa by Walid Mustafa , 22, Ireland

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1 solution

Walid Mustafa
Jan 3, 2015

My first solution (YAY!):

S u p p o s e t h a t t h e c l a s s t r i e d e v e r y c o n f i g u r a t i o n . B o y i a n d g i r l j w o u l d s t a n d n e x t t o e a c h o t h e r i n 2 d i f f e r e n t o r d e r s , i n 19 d i f f e r e n t p o s i t i o n s , 18 ! t i m e s e a c h . S u m m i n g o v e r a l l i , j g i v e s 7 ⋅ 13 ⋅ 2 ⋅ 19 ⋅ 18 ! = 91 10 ⋅ 20 ! , s o t h e a v e r a g e v a l u e o f S i s 9 1 10 ( A ) . Suppose\quad that\quad the\quad class\quad tried\quad every\quad configuration.\\ Boy\quad i\quad and\quad girl\quad j\quad would\quad stand\quad next\quad to\quad each\quad other\\ in\quad 2\quad different\quad orders,\quad in\quad 19\quad different\quad positions,\\ 18!\quad times\quad each.\quad Summing\quad overall\quad i,\quad j\quad gives\quad \\ 7\quad \cdot \quad 13\quad \cdot \quad 2\quad \cdot \quad 19\quad \cdot \quad 18!\quad =\quad \frac { 91 }{ 10 } \quad \cdot \quad 20!,\quad so\quad the\quad \\ average\quad value\quad of\quad S\quad is\quad \boxed { 9\frac { 1 }{ 10 } (A) } .\quad

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