My Math Forum Arranging people in a line

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 August 18th, 2017, 10:59 AM #1 Member   Joined: Apr 2014 From: Greece Posts: 58 Thanks: 0 Arranging people in a line Can somebody help me understand this? We have $\displaystyle b$ boys and $\displaystyle g$ girls and we arrange them in a line in random order. What is the probability that there won't be consecutive girls in the line? The answer was given to me and it's $\displaystyle \frac{b!\binom{b+1}{g}g!}{(b+g)!}$ I understand that $\displaystyle b!$ and $\displaystyle g!$ are the possible positions for the boys and the girls and that $\displaystyle (b+g)!$ are the possible lines that can be created but I don't understand what the $\displaystyle \binom{b+1}{g}$ stands for.. Can any one explain??
 August 18th, 2017, 11:57 AM #2 Senior Member     Joined: Feb 2010 Posts: 674 Thanks: 127 As you said $\displaystyle (b+g)!$ is the number of ways to line up everyone. For the numerator ... line up the boys in $\displaystyle b!$ ways. Now spread them out so there is room for one person in between ... like this: _b_b_b_b_ ... _b_b_ If there are $\displaystyle b$ boys then there will be $\displaystyle b+1$ spaces (including before the first and after the last) Choose $\displaystyle g$ spots to place the girls in $\displaystyle \binom{g}{b+1}$ ways. (I think you had this upside down.) Now line up the girls in $\displaystyle g!$ ways and insert them into the chosen spaces. So the numerator is as you say it is ... line up the boys, figure out where to place the girls, line up the girls and place them. Last edited by mrtwhs; August 18th, 2017 at 11:59 AM. Reason: addition
 August 18th, 2017, 03:36 PM #3 Senior Member     Joined: Feb 2010 Posts: 674 Thanks: 127 Well I goofed big time. You were correct on the binomial coefficient. It is $\displaystyle \binom{b+1}{g}$. I'll go and stand in the corner.
August 19th, 2017, 07:36 AM   #4
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Quote:
 Originally Posted by Vaki The answer was given to me and it's $\displaystyle \frac{b!\binom{b+1}{g}g!}{(b+g)!}$
$\displaystyle b!\binom{b+1}{g}g!$

where,
$\displaystyle \binom {b+1}{g} = (b+1)\hspace{2mm} combination \hspace{2mm}g$

August 19th, 2017, 07:41 AM   #5
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 Originally Posted by MATHEMATICIAN I think the answer is $\displaystyle b!\binom{b+1}{g}g!$ where, $\displaystyle \binom {b+1}{g} = (b+1)\hspace{2mm} combination \hspace{2mm}g$
That's the number of ways to line the people up such that no two girls are adjacent. To get the probability, you'd divide that through by the total number of ways to line the people up (which is $(b+g)!$).

 August 19th, 2017, 07:49 AM #6 Math Team     Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 878 Thanks: 60 Math Focus: सामान्य गणित Explanation: First, let us take "b" boys and arrange them in b! ways. After arranging them we can insert "g" girls in "b+1" spaces. (B let) number of arrangements of "b" boys = b! (G let) number of arrangements of "g" girls in "b+1" spaces = (b+1)!/[(b+1)-g]! So, total number of arrangements = B × G $\displaystyle =b!×\frac {(b+1)!}{[(b+1)-g]!}$ $\displaystyle =b!×g!×\frac {(b+1)!}{[(b+1)-g]!×g!}$ $\displaystyle =b!×g!×\binom {b+1}{g}$ NOTE: b > g
August 19th, 2017, 07:51 AM   #7
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 Originally Posted by cjem That's the number of ways to line the people up such that no two girls are adjacent. To get the probability, you'd divide that through by the total number of ways to line the people up (which is $(b+g)!$).
I did not see the word probability

August 19th, 2017, 07:58 AM   #8
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Quote:
 Originally Posted by MATHEMATICIAN Explanation: First, let us take "b" boys and arrange them in b! ways. After arranging them we can insert "g" girls in "b+1" spaces. (B let) number of arrangements of "b" boys = b! (G let) number of arrangements of "g" girls in "b+1" spaces = (b+1)!/[(b+1)-g]! So, total number of arrangements = B × G $\displaystyle =b!×\frac {(b+1)!}{[(b+1)-g]!}$ $\displaystyle =b!×g!×\frac {(b+1)!}{[(b+1)-g]!×g!}$ $\displaystyle =b!×g!×\binom {b+1}{g}$ NOTE: b > g
I agree with everything you've said (for calculating the number of ways to line them up in this way), except that $b > g$ is too strict of a requirement. So long as $b + 1 \geq g$, your explanation works and the probability is greater than zero. It's only when $b + 1 < g$ that the probability becomes zero, and it stops making sense to talk of $b+1 \choose g$.

August 19th, 2017, 08:09 AM   #9
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Quote:
 Originally Posted by cjem I agree with everything you've said (for calculating the number of ways to line them up in this way), except that $b > g$ is too strict of a requirement. So long as $b + 1 \geq g$, your explanation works and the probability is greater than zero. It's only when $b + 1 < g$ that the probability becomes zero, and it stops making sense to talk of $b+1 \choose g$.
Correction: $\displaystyle b \geqslant g$

August 19th, 2017, 08:15 AM   #10
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Quote:
 Originally Posted by MATHEMATICIAN Correction: $\displaystyle b \geqslant g$
Correction: $b+1 \geq g$

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