# The proof of binomial identity $\binom{j+r-1}{j}=(-1)^j \binom{-r}{j}$?

I'm trying to understand the following proof for $\binom{j+r-1}{j}=(-1)^j \binom{-r}{j}$.

\begin{align} \binom{j+r-1}{j}&=\frac{(j+r-1)(j+r-2) \cdots r}{j!}\\ &=(-1)^j \frac{(-r-(j-1))(-r-(j-2)) \cdots (-r)}{j!} \\&=(-1)^j \frac{(-r)(-r-1) \cdots (-r-(j-1))}{j!} \\&=(-1)^j \binom{-r}{j} \end{align}

I think I understand everything else than the second equation.

The first equation merely uses the definition of a binomial coefficient.
The second one seems to change the signs of the terms, but why does $(-1)$ have to have $j$th power?
The third equation merely reverses the nominator terms in order to see the pattern that can be written as a binomial coefficient in the fourth equation.

So is this correct and can you explain the second equation?

• you cange sign in any of the $j$ parts of the product, given an overall $(-1)^j$. the other stuff is nicely explained, well done! – tired Nov 30 '16 at 11:01
• @tired Do you mean that if one multiplied by $(-1)$ then one couldn't control what sign actually results, because that would depend on what sign $(j+r-1)$, $(j+r-2)$...$r$ each have? But if one multiplies by $(-1)^j$ then one switches the sign of only one term, thus certainly flipping the sign of the entire nominator? – mavavilj Nov 30 '16 at 11:07
• I don't really understand why the power has to be exactly $j$. – mavavilj Nov 30 '16 at 11:22
• try a small example. maybe $j=4$ – tired Nov 30 '16 at 11:27
• @tried But couldn't one try e.g. 1*2*3 and then if one needs to flip the sign it would mean to multiply by (-1)^3, i.e. 1*2*3=(-1)^3(-1)(-2)(-3)? Yea it seems to work and the proof is probably by induction. – mavavilj Nov 30 '16 at 14:48

Your explanation is correct. Note the binomial coefficient \begin{align*} \binom{n}{j}=\frac{n(n-1)\cdots(n-(j-1))}{j!} \end{align*} has according to the $j$ factors of $j!=j(j-1)\cdots 3\cdot 2\cdot 1$ also $j$ factors in the numerator.
We obtain \begin{align*} \binom{j+r-1}{j}&=\binom{r-1+j}{j}\\ &=\frac{(r-1+\color{blue}{j})(r-1+\color{blue}{(j-1)})\cdots(r-1+\color{blue}{(2)})(r-1+\color{blue}{(1)})}{j!}\\ &=\frac{(r-1+j)(r-2+j)\cdots(r+1)r}{j!}\\ &=(-1)^j\frac{(-r+1-j)(-r+2-j)\cdots(-r-1)(-r)}{j!}\tag{1}\\ \end{align*} In (1) we factor out $(-1)$ from each of the $j$ factors of the numerator giving $(-1)^j$.
• I like your explanation of "factor out $(-1)$ from each of the $j$ factors". This makes it very clear. – mavavilj Nov 30 '16 at 14:48
$$\binom 75=\frac{7\cdot6\cdot5\cdot4\cdot3}{5!}=(-1)^5\frac{(-3)(-4)(-5)(-6)(-7)}{5!}$$
Notice now that $(-3)(-4)(-5)(-6)(-7)=-3(-3-1)(-3-2)(-3-3)(-3-4)$ which is the numerator of $\binom{-3}{ 5}$. You can see clearly in this example that $$\binom 75=(-1)^5\binom{-3}{ 5}$$ In this case $j=5,\space r=3$ so $j+r-1=7$