Problem with binomial coefficients

Question

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?

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. ain't a mathematician, Oct 03 '10 at 10:37
(-2)! is not -2; it's not even defined. http://en.wikipedia.org/wiki/Factorial#Non-extendability_to_negative_integers — , Oct 03 '10 at 10:37
What is your definition of the binomial coefficient? That's really what your question comes down to. — Mike Spivey, Nov 30 '11 at 04:53
In other words, we don't want to say that $(-n)! = \infty$ but we have no problem saying that $\dfrac{1}{(-n)!} = 0$. — Steven Alexis Gregory, Jul 17 '15 at 01:52

score 3 · Accepted Answer · answered Oct 03 '10 at 11:30

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.

score 2 · Answer 2 · answered Oct 03 '10 at 19:51

2

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

answered Oct 03 '10 at 19:51

Bill Dubuque

272,048

score 2 · Answer 3 · answered Oct 03 '10 at 21:05

$(-2)!$ 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.

score 1 · Answer 4 · answered May 08 '13 at 07:35

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.

Problem with binomial coefficients

4 Answers4