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August 19th, 2017, 08:18 AM  #11 
Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 878 Thanks: 60 Math Focus: सामान्य गणित  
August 19th, 2017, 08:21 AM  #12 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 188 Thanks: 57 Math Focus: Algebraic Number Theory, Arithmetic Geometry  
August 19th, 2017, 08:26 AM  #13 
Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 878 Thanks: 60 Math Focus: सामान्य गणित  
August 19th, 2017, 12:07 PM  #14 
Senior Member Joined: Feb 2010 Posts: 674 Thanks: 127 
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 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 188 Thanks: 57 Math Focus: Algebraic Number Theory, Arithmetic Geometry  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. 

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