Cierra Castillo

2022-06-29

Binomial coefficient inequality equation proofs
Prove the statement:
If n is a positive integer, then $1=\left(\genfrac{}{}{0}{}{n}{0}\right)<\left(\genfrac{}{}{0}{}{n}{1}\right)<...<\left(\genfrac{}{}{0}{}{n}{⌊n/2⌋}\right)=\left(\genfrac{}{}{0}{}{n}{⌈n/2⌉}\right)>...>\left(\genfrac{}{}{0}{}{n}{n-1}\right)>\left(\genfrac{}{}{0}{}{n}{n}\right)=1$
I tried to read up the falling/rising factorial notation from internet sources, but I cannot find how to write up the first step.
The options are as follows (We have to put it in the right order, one option is incorrect):
- If $k<=n/2$, then $k/\left(n—k+1\right)$, so the "greater than" signs are correct. Similarly, if $k>n/2$, then $k/\left(n-k+1\right)$, so the "less than" signs are correct.
- If $k<=n/2$, then $k/\left(n—k+1\right)$, so the "less than" signs are correct. Similarly, if $k>n/2$, then $k/\left(n-k+1\right)$, so the "greater than" signs are correct.
- Since $⌊n/2⌋+⌈n/2⌉=n$, the equalities at the ends are clear.
- The equalities at the ends are clear. Using the factorial formulae for computing binomial coefficients, we see that $C\left(n,k—1\right)=\left(k/\left(n-k+1\right)\right)C\left(n,k\right)$
If we consider a sample k as $k=2...k=5$ and we consider $n=7$.
Then $\left(\genfrac{}{}{0}{}{n}{3}\right)=\left(\genfrac{}{}{0}{}{n}{⌊n/2⌋}\right)$ because the floor value of n/2 is 3. But the same holds good on the right side of the equation with greater than signs with $\left(\genfrac{}{}{0}{}{n}{4}\right)=\left(\genfrac{}{}{0}{}{n}{⌈n/2⌉}\right)$ because the ceiling value of n/2 is 4.
It is confusing to understand which of the steps is correct between the ${1}^{st}$ and ${2}^{nd}$ provided in the option list. Also what is the first step for solving this, the problem definition or something similar does not exist in the text provided.

### Answer & Explanation

Caiden Barrett

Step 1
I got the answer wrong. Here is the answer :
Using the factorial formulae for computing binomial coefficients, we see that $C\left(n,k–1\right)=\left(k/\left(n–k+1\right)\right)C\left(n,k\right)$
If $k\le n/2,then,k/\left(n–k+1\right)<1$, so the "less than" signs are correct. Similarly, if $k>n/2,then,k/\left(n–k+1\right)>1$, so the "greater than" signs are correct.
Step 2
Since
Thus, the middle equality is clear.
The equalities at the ends are clear.

Joshua Foley

Step 1
$\begin{array}{rl}\left(\genfrac{}{}{0}{}{n}{k+1}\right)-\left(\genfrac{}{}{0}{}{n}{k}\right)& =\frac{n!}{\left(n-k-1\right)!\left(k+1\right)!}-\frac{n!}{\left(n-k\right)!k!}\\ & =\frac{n!}{\left(n-k-1\right)!\left(k-1\right)!}\cdot \left(\frac{1}{k+1}-\frac{1}{n-k}\right)\end{array}$
Step 2
Thus .

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