showing that $\binom{m^{3}}{m+1}= \binom{m^{3}}{m}\cdot(m^2-m)$

Question

I have a problem.

$$\binom{m^{3}}{m+1}= \binom{m^{3}}{m}\cdot(m^2-m)$$

I don't know what to do to make the right side equal with left. Everytime I do that it doesn't work. Step by step explanation please!

Right side: (m^3!)/(m!(m^3-m))! * (m^2-m)

Answer 1

You could break down the binomial coefficients into factorials. So you are trying to show:

$$\frac{(m^3)!}{(m+1)!(m^3-m-1)!}=\frac{(m^3)!}{m!(m^3-m)!}\cdot(m^2-m)$$

If you divide on both sides by $m^3-m$, you make the second factor in the left denominator match the second factor in the right denominator:

$$\frac{(m^3)!}{(m+1)!(m^3-m-1)!(m^3-m)}=\frac{(m^3)!}{m!(m^3-m)!}\cdot\frac{(m^2-m)}{m^3-m}$$ $$\frac{(m^3)!}{(m+1)!(m^3-m)!}=\frac{(m^3)!}{m!(m^3-m)!}\cdot\frac{(m^2-m)}{m^3-m}$$

If you multiply on each side by $m+1$, you do the same with the first factor of the denominator:

$$\frac{(m+1)(m^3)!}{(m+1)!(m^3-m)!}=\frac{(m^3)!}{m!(m^3-m)!}\cdot\frac{(m^2-m)(m+1)}{m^3-m}$$ $$\frac{(m^3)!}{m!(m^3-m)!}=\frac{(m^3)!}{m!(m^3-m)!}\cdot\frac{(m^2-m)(m+1)}{m^3-m}$$

Now what does that fraction on the far right simplify to?

Note that all of the equations here are equivalent, since we only ever multiplied/divided by nonzero quantities.

Answer 2

Consider $\binom{a}{m}$ and $\binom{a}{m+1}$. By definition, we have that $$ \binom{a}{m+1} = \frac{a\times (a-1)\times \cdots\times (a-m)}{(m+1)\times m\times\cdots 1}. $$ Similarly, $$ \binom{a}{m} = \frac{a\times (a-1)\times\cdots\times (a-m+1)}{m\times (m-1)\times \cdots\times 1}. $$ Compare the two expressions. We have that $$ \binom{a}{m+1} = \frac{a-m}{m+1}\binom{a}{m}. $$ Now take $a=m^3$ you get what you want.

Answer 3

Expand the wrong side: $C_{m^3}^{m+1}=\frac{(m^3)!}{(m+1)! \cdot (m^3-m-1)!}$

Do the same with right side: $C_{m^3} ^{m} \cdot (m^2-m)=\frac{(m^3)!}{m! \cdot (m^3-m)!} \cdot (m^2-m)=C_{m^3}^ {m+1}\cdot \frac{m+1}{m^3-m}\cdot (m^2-m)=C_{m^3}^ {m+1}(\therefore)$

There you go: right is left. Or wrong. Same thing.

Answer 4

You can't think of it combinatorically. ${m^3}\choose{m+1}$ is the number of ways of choosing $m+1$ objects from the set of $m^3$ objects and the order doesn't matter. On the other hand, it's the same as first choosing $m$ objects, which is ${m^3}\choose{m}$ and then choosing one object from the remaining ones, which is $m^3-m$. But you counted the same set $m+1$ times, so you divide ${m^3}\choose{m}$$(m^3-m)$ by $m+1$, which simplifies to ${m^3}\choose{m}$$(m^2-m)$, the right side of the given identity.