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?
 A: 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.
A: By definition $\rm\binom{n}k$ is the coefficient of $\rm x^k$ in $\rm (1+x)^n$ so it is $0$ for $\rm k > n\:$.
A: $(-2)!$ 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. 
A: 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.
