Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
4  
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. –  J. M. Oct 3 '10 at 10:37
1  
(-2)! is not -2; it's not even defined. en.wikipedia.org/wiki/… –  Rahul Oct 3 '10 at 10:37
1  
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

4 Answers 4

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.

share|improve this answer

$(-2)!$ 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.

share|improve this answer

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

share|improve this answer
up vote 3 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.