How do you solve the equation:

$$ \binom{n}{3}=4\binom{n}{2} $$

I've tried multiple times I'm just not sure what I'm supposed to do to be honest.

What I did try doing was the combinatorial theorem, and re-arranging to get n, but I didnt get a solution. So I'm wondering how I would go about solving for n.

  • $\begingroup$ do you mean $\binom{n}{3}=4\binom{n}{2}$? $\endgroup$ – Cornman Nov 26 '17 at 17:29
  • $\begingroup$ yea thats it thanks $\endgroup$ – John Nov 26 '17 at 17:33
  • $\begingroup$ Isn't it straightforward by following the definition? $\endgroup$ – Von Neumann Nov 26 '17 at 17:52

You can just write the definitions on each side and solve for $n$:

$$\dfrac {n!}{3!(n-3)!}=4\dfrac {n!}{2!(n-3)!}$$

$$\iff \dfrac {1}{6(n-3)!}=\dfrac {2}{(n-2)!}$$

$$\iff \dfrac {(n-2)!}{(n-3)!}=12$$

$$\iff (n-2)=12$$

$$\iff n=14$$

  • $\begingroup$ It's common to define $\binom{n}{k}$ to be zero if $n<k$. If you follow that convention, then 0 and 1 are also solutions. $\endgroup$ – Rick Decker Nov 26 '17 at 20:21

If you write out these expressions in terms of their definitions, you are greeted with simply a polynomial equation:

$${n\choose3} = 4{n\choose2} \implies \frac{n(n-1)(n-2)}{3!}=\frac{4n(n-1)}{2!}.$$

From here, it should be very easy to extract such a value for $n$.

  • $\begingroup$ And in fact for $n \ge 2$ you can cancel $n(n - 1)$ and the resulting equation is linear in $n$! $\endgroup$ – Robert Lewis Nov 26 '17 at 17:35

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