Inequality of binomial coefficients

Question

Let $n$ be a natural number. we need to proove that: $$\forall{1\leq k\leq\tfrac{n}2}, \text{we have } {n \choose k-1} < {n \choose k} $$

I tried to do the following with induction:

We set $n$ as a constant and doing induction on $k$: if $ k=1 $: ${n \choose 1} > {n \choose 0}$ and it can be shown by writing the exlpicit ration of the ${i\choose j} sign $ . Assuming that for every $k$ as it was set, the equation above is right, we will try to proove with induction that for $k+1$ we get: $$\forall{1\leq k+1\leq\tfrac{n}{2}}, \text{we have}{n\choose k} < {n \choose k+1}$$

But at this point, I am stuck, because I can't see how can I apply the induction assumption.

I also tried to use pascal identity but I can't see how it fits in.

yes, you are right. I had just updated that one. I managed to proove the (example) base case but I am stuck in the Induction step itself. — Tsviel Zaikman, May 31 '21 at 10:36
Does this answer your question? Proving If $k \le \left\lfloor \frac{n}{2} \right\rfloor$ then $\binom{n}{k-1} < \binom{n}{k}$ — Martin R, May 31 '21 at 10:58

score 3 · Answer 1 · answered May 31 '21 at 10:44

We want to show that

$\binom{n}{k-1}<\binom{n}{k}$ for all $1\leq k\leq n/2$.

With a direct calculation, we get:

$\binom{n}{k-1}=\frac{n!}{(n-k+1)!(k-1)!}\stackrel{?}{<}\frac{n!}{k!(n-k)!}=\binom{n}{k}$

We can cancel $n!$ and the inequality in question is equivalent to

$(n-k+1)!(k-1)!>(n-k)!k!$

Dividing $(n-k)!$ and $(k-1)!$, we get $n-k+1>k\Leftrightarrow n+1>2k$, as $k\leq n/2$

we have $n+1>n=2\cdot \frac{n}2\geq 2k$.

Inequality of binomial coefficients

1 Answers1