
Probability and Statistics Basic Probability and Statistics Math Forum 
 LinkBack  Thread Tools  Display Modes 
August 19th, 2017, 08:18 AM  #11 
Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 831 Thanks: 60 Math Focus: सामान्य गणित  
August 19th, 2017, 08:21 AM  #12 
Member Joined: Aug 2017 From: United Kingdom Posts: 90 Thanks: 26  
August 19th, 2017, 08:26 AM  #13 
Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 831 Thanks: 60 Math Focus: सामान्य गणित  
August 19th, 2017, 12:07 PM  #14 
Senior Member Joined: Feb 2010 Posts: 627 Thanks: 98 
The condition $\displaystyle b+1 \geq g$ is not needed. Suppose $\displaystyle b=5$ and $\displaystyle g=7$. The the binomial coefficient is $\displaystyle \binom{6}{7}$ which equals zero and that is the correct answer.

August 19th, 2017, 12:13 PM  #15 
Member Joined: Aug 2017 From: United Kingdom Posts: 90 Thanks: 26  I guess this is just a matter of convention. I've only ever seen $n \choose r$ defined for $n \geq r$ but, yes, if you define it to be zero when $n < r$, it will make the formula hold for all integers $b,g \geq 0$.
Last edited by cjem; August 19th, 2017 at 12:32 PM. 

Tags 
arranging, line, people 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Arranging addends  Loren  Number Theory  6  May 9th, 2015 11:41 PM 
Rearranging formula's  kiwiginga  Algebra  2  May 9th, 2015 09:19 PM 
Help rearranging formula  danw76  Algebra  1  October 27th, 2014 01:25 PM 
How many ways can nine people line up?  RedBarchetta  Probability and Statistics  8  June 29th, 2014 09:15 PM 
Arranging quadratics  dthomas86  Algebra  2  June 7th, 2012 05:29 AM 