1

So, if we're given something like this: $$\binom{n}{k}:\binom{n+1}{k}:\binom{n+1}{k+1}=3:4:8$$ How do I rewrite this so I can manipulate it?

Edit: Is there a general procedure for n variables?

A6SE
  • 1,645
  • 1
    We have ${n\choose k} = \frac{n!}{(n-k)! k!} = \frac{n+1-k}{n+1} \frac{(n+1)!}{(k+1)!(n+1-k)!} = \frac{n+1-k}{n+1} {n+1\choose k}$ and you can derive similar expressions for the other fractions to get a simple equation system. – Winther Nov 07 '15 at 14:39
  • Thank you! One more question, what happens if we have only two binomial coefficients in a proportion and we want to get n and k? There seems to be lack of equations to solve such case. – A6SE Nov 07 '15 at 20:11
  • 1
    Yes. One gets a relation between $n$ and $k$. For example the $3$:$4$ fraction is satisfied for all integers $n,k$ satisfying $4k = n+1$. – Winther Nov 07 '15 at 20:17

1 Answers1

1

Note, that the equality of proportions \begin{align*} a:b:c=x:y:z \end{align*} is a compact notation for \begin{align*} \frac{a}{b}=\frac{x}{y},\qquad\frac{a}{c}=\frac{x}{z},\qquad\frac{b}{c}=\frac{y}{z}\tag{1} \end{align*} Each equality in (1) can be derived from the other two.

So, in order to check if we can solve \begin{align*} \binom{n}{k}:\binom{n+1}{k}:\binom{n+1}{k+1}=3:4:8 \end{align*} we take two proportions and calculate $n$ and $k$.

We find by solving \begin{align*} \frac{\binom{n}{k}}{\binom{n+1}{k}}=\frac{3}{4}\qquad \frac{\binom{n+1}{k}}{\binom{n+1}{k+1}}=\frac{4}{8} \end{align*}

the solutions $n=7,k=2$ and can finally conclude \begin{align*} \binom{7}{2}:\binom{8}{2}:\binom{8}{3}=21:28:56=3:4:8 \end{align*}

Markus Scheuer
  • 108,315