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Above is a part of calculation involving the binomial coefficient. I understand the steps. What we have done is that we added and subtracted the same terms (those drawn in a red box) so that nothing changed by this step. But these terms don't exist in the definition of a binomial coefficient.. So is it logical to add or subtract a value that doesn't exist?! I hope that I were clear in presenting what was unclear to me.

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    $\begingroup$ These coefficients are defined as zeroes. $\endgroup$ – Yves Daoust Dec 27 '17 at 17:04
  • $\begingroup$ Well, you can't just throw them away. You must either define or demonstrate that they are equal to 0. If you want to take it literally: There are 0 ways to choose a negative number from a positive number, and there are 0 ways to choose more numbers than are offered. That sounds like a joke but .... dang if it isn't true. Or, for convenience we can simply define them as 0 as Yves suggest. Or we can define them recursively as celtschk suggests at the end of his answer. $\endgroup$ – fleablood Dec 27 '17 at 18:11

I guess you are used to the definition of the binomial coefficients as $$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$ However if you cancel out the $(n-k)!$, you get $$\binom{n}{k} = \frac{n(n-1)(n-2)\cdots(n-k+1)}{k!}$$ And if you write it in this way, you see that there is no problem with having arbitrary values for $n$. For example, you have $$\binom{\pi}{3} = \frac{\pi(\pi-1)(\pi-2)}{6}$$ Or in your case, $$\binom{n-1}{n} = \frac{(n-1)(n-2)\cdots ((n-1)-n+1)}{n!}$$ Now notice that the last factor in the numerator is actually zero; therefore the whole binomial coefficient vanishes.

Another way to get to the same result is to notice that there is a recursion formula, $$\binom{n}{k+1} = \frac{n-k}{k+1}\binom{n}{k}$$ which is easily checked by inserting the original definition. By assuming this recursion formula to hold further, we then get immediately $$\binom{n-1}{n} = \binom{n-1}{(n-1)+1} = \frac{(n-1)-(n-1)}{(n-1)+1}\binom{n-1}{n-1} = 0$$

OK, that takes care of the first term, but what about the second? Even with this definition, you'd still get $(-1)!$ in the denominator. There are again at least two ways to tackle this.

The first one is to observe that $\binom{n}{k} = \binom{n}{n-k}$ and just assume that this also holds also outside the original range. Then you get $$\binom{n-1}{-1} = \binom{n-1}{(n-1)-(-1)} = 0$$

A second way is to solve the recursion formula from above for $\binom{n}{k}$: $$\binom{n}{k} = \frac{k+1}{n-k}\binom{n}{k+1}$$ Then we get immediately $$\binom{n-1}{-1} = \frac{-1+1}{n-1-(-1)}\binom{n-1}{-1+1} = 0$$

In any case, you get that those extra terms added are well defined and have the value $0$.


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