1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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! $\endgroup$ – tired Nov 30 '16 at 11:01
  • $\begingroup$ @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? $\endgroup$ – mavavilj Nov 30 '16 at 11:07
  • $\begingroup$ I don't really understand why the power has to be exactly $j$. $\endgroup$ – mavavilj Nov 30 '16 at 11:22
  • $\begingroup$ try a small example. maybe $j=4$ $\endgroup$ – tired Nov 30 '16 at 11:27
  • $\begingroup$ @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. $\endgroup$ – mavavilj Nov 30 '16 at 14:48
1
$\begingroup$

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 $(-1)^j$.

$\endgroup$
  • $\begingroup$ I like your explanation of "factor out $(-1)$ from each of the $j$ factors". This makes it very clear. $\endgroup$ – mavavilj Nov 30 '16 at 14:48
  • $\begingroup$ @mavavilj: You're welcome! Good to see the answer is useful. :-) $\endgroup$ – Markus Scheuer Nov 30 '16 at 14:52
0
$\begingroup$

It is just fitting introduction of signes and inversion of a decreasing and consecutive (finite) sequence of factors. Look at the following example:

$$\binom 75=\frac{7\cdot6\cdot5\cdot4\cdot3}{5!}=(-1)^5\frac{(-3)(-4)(-5)(-6)(-7)}{5!}$$

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 $j+r-1=7$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.