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Comparing the formula for regular binomial expansion (n>1):

$(a+b)^n=a^n + \binom{n}1a^{n-1}b + \binom{n}2a^{n-2}b^2 +...$

to binomial expansion for negative indices, (n<1):

$(1+x)^n= 1 + nx + \dfrac{n(n-1)x^2}{2!} +...$

Does this imply that $\binom{-1}2= \dfrac{-1!}{2!(-1-2)!}=\dfrac{-1.-2}{2!} $?

Confused by the negative factorial...

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(Backstory: haven't done maths in a very long time)

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For some reason that I cannot really understand (except if it is purely aesthetic) many people want to define binomial coefficients by $\binom nk=\frac{n!}{k!(n-k)!}$. But that formula is neither very efficient, nor very general (it is only valid for integers $0\leq k\leq n$). Of course binomial coefficients should be defined combinatorially, but if one looks at the basic argument that leads to the given "closed form", then it actually directly gives $$ \binom nk = \frac{n(n-1)\ldots(n-k+1)}{k!} $$ (namely: choose a $k$-element subset by choosing the first element in $n$ ways, the second in $n-1$ ways, ... the last in $n-k+1$ ways, and divide that product by $k!$ to compensate for the different orderings for which each subset is obtained). Throwing in $(n-k)!$ in numerator and denominator gives the "three factorials" form.

The displayed formula is a much better one when it comes to generalising; as long as $k\in\Bbb N$, one can allow $n$ to be anything, including a negative integer, fraction, or even a formal indeterminate (in which case one obtains a polynomial of degree$~k$ with rational coefficients, taking integer values at all integers). It is sometimes useful to extend the definition so obtained by stipulating $\binom nk=0$ whenever $k$ is a negative integer, regardless of the value of$~n$ (it is more generally often useful to take $\frac1{k!}=0$ for negative integer$~k$). One reason that the generalisation is useful is the binomial formula $$ (1+X)^\alpha = \sum_{k\in\Bbb N}\binom\alpha kX^k $$ that is valid as an identity of formal power series for arbitrary values of$~\alpha$, including negative integers and fractions. (Substituting $z$ for $X$ gives a converging series as right hand side whenever $|z|<1$.)

To link this with your question, it means $\binom{-1}2=\frac{-1\times-2}{2!}=1$ is correct, just skip the meaningless intermediate formula.

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No, the factorial function isn't defined for negative integers, and the choose function is usually either said to be undefined, or equal zero, when some argument is negative. Although such manipulations sometimes lead to interesting ideas on how one might extend things, the moral here is not to apply theorems or functions outside thier scope.

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    $\begingroup$ I don't agree with "usually either said to be undefined, or equal zero". The extension to arbitrary upper indices mentioned in my answer is both common and very useful. You can find it used in many questions on this site, for instance the ones linked in my comment to the question. See also binomial series. $\endgroup$ – Marc van Leeuwen Jan 21 '15 at 8:39

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