August 18th, 2017, 10:59 AM  #1 
Member Joined: Apr 2014 From: Greece Posts: 52 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: 627 Thanks: 98 
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: 627 Thanks: 98 
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 
Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 831 Thanks: 60 Math Focus: सामान्य गणित  
August 19th, 2017, 07:41 AM  #5 
Member Joined: Aug 2017 From: United Kingdom Posts: 90 Thanks: 26  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: 831 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 
Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 831 Thanks: 60 Math Focus: सामान्य गणित  
August 19th, 2017, 07:58 AM  #8  
Member Joined: Aug 2017 From: United Kingdom Posts: 90 Thanks: 26  Quote:
 
August 19th, 2017, 08:09 AM  #9  
Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 831 Thanks: 60 Math Focus: सामान्य गणित  Quote:
 
August 19th, 2017, 08:15 AM  #10 
Member Joined: Aug 2017 From: United Kingdom Posts: 90 Thanks: 26  

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