# Problem with binomial coefficients

I have a problem with the binomial coefficient $\binom{5}{7}$. I know that the solution is zero, but I have problems to reproduce that:

${\displaystyle \binom{5}{7}=\frac{5!}{7!\times(5-7)!}=\frac{5!}{7!\times(-2)!}=\frac{120}{5040\times-2}=\frac{120}{-10080}=-\frac{1}{84}}$

Where is my mistake?

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It's a bit of a cheat, but $\binom{m}{n}$ for $m < n$ is 0 because the reciprocal gamma function (and thus the reciprocal factorial) is zero at the negative integers. –  Guess who it is. Oct 3 '10 at 10:37
(-2)! is not -2; it's not even defined. en.wikipedia.org/wiki/… –  Rahul Oct 3 '10 at 10:37
Are you sure that $(-2)! = -2$? –  a.r. Oct 3 '10 at 10:38
Yes, now I see. Thank you. –  Sven Walter Oct 3 '10 at 11:31
What is your definition of the binomial coefficient? That's really what your question comes down to. –  Mike Spivey Nov 30 '11 at 4:53

Although not as formal, one by relying on only a combinatoric definition of the binomial coefficient we can find that it is zero straight away. Consider that:

$\binom{m}{k}$ is the amount of combinations of k elements from m numbered set.

It is obvious that we cannot select more than what we have, so if m < k, then the answer is already zero because for example in this case, if we have 5 apples, then it is impossible to select 7 apples from the 5, hence there are zero combinations.

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$(-2)!$ is actually infinite. A more palatable way to phrase that, perhaps, is in terms of the reciprocal factorial: $1/(-2)! = 0$. We only need the recurrence relation $n! = n(n-1)!$, or in terms of reciprocal factorials: $$\frac{1}{(n-1)!} = n\cdot\frac{1}{n!}.$$ That means $\frac{1}{(-2)!} = \frac{0\cdot (-1)}{0!} = 0$. Then $\binom{5}{7} = \frac{5!}{7!}\cdot \frac{1}{(-2)!} = 0$, QED.

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By definition $\rm\binom{n}k$ is the coefficient of $\rm x^k$ in $\rm (1+x)^n$ so it is $0$ for $\rm k > n\:$.

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Yes, now I see the problem.

First, (-2)! really isn't defined. And I can't use the factorial method if $n\notin\mathbb{N}$. So I have to go these way:

${\displaystyle \binom{5}{7}=\frac{5\times4\times3\times2\times1\times0\times-1}{7!}=\frac{0}{7!}=0}$

Thus, if $k>n$ the solution will always be zero, because the numerator has always the factor zero.

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