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August 19th, 2017, 08:18 AM   #11
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Quote:
 Originally Posted by cjem correction: $b+1 \geq g$
gbgbgbgbg ok

August 19th, 2017, 08:21 AM   #12
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Quote:
 Originally Posted by MATHEMATICIAN gbgbgbgbg ok
Yeah, exactly. It's just a small point, though, and I was nitpicking :P You got the harder part correct.

August 19th, 2017, 08:26 AM   #13
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Quote:
 Originally Posted by cjem Yeah, exactly. It's just a small point, though, and I was nitpicking :P You got the harder part correct.
i am not good at thinking

 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
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Quote:
 Originally Posted by mrtwhs 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.
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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