1
$\begingroup$

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?

  • 0
    In other words, we don't want to say that $(-n)! = \infty$ but we have no problem saying that $\dfrac{1}{(-n)!} = 0$.2015-07-17

4 Answers 4

3

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.

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\:.

2

$(-2)!$ 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.

1

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.