Prove the statement:

If n is a positive integer, then $1={\textstyle (}\genfrac{}{}{0ex}{}{n}{0}{\textstyle )}<{\textstyle (}\genfrac{}{}{0ex}{}{n}{1}{\textstyle )}<...<{\textstyle (}\genfrac{}{}{0ex}{}{n}{\lfloor n/2\rfloor}{\textstyle )}={\textstyle (}\genfrac{}{}{0ex}{}{n}{\lceil n/2\rceil}{\textstyle )}>...>{\textstyle (}\genfrac{}{}{0ex}{}{n}{n-1}{\textstyle )}>{\textstyle (}\genfrac{}{}{0ex}{}{n}{n}{\textstyle )}=1$

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 $k/(n\u2014k+1)$, so the "greater than" signs are correct. Similarly, if $k>n/2$, then $k/(n-k+1)$, so the "less than" signs are correct.

- If $k<=n/2$, then $k/(n\u2014k+1)$, so the "less than" signs are correct. Similarly, if $k>n/2$, then $k/(n-k+1)$, so the "greater than" signs are correct.

- Since $\lfloor n/2\rfloor +\lceil n/2\rceil =n$, 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 $C(n,k\u20141)=(k/(n-k+1))C(n,k)$

If we consider a sample k as $k=2...k=5$ and we consider $n=7$.

Then $(}\genfrac{}{}{0ex}{}{n}{3}{\textstyle )}={\textstyle (}\genfrac{}{}{0ex}{}{n}{\lfloor n/2\rfloor}{\textstyle )$ 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 $(}\genfrac{}{}{0ex}{}{n}{4}{\textstyle )}={\textstyle (}\genfrac{}{}{0ex}{}{n}{\lceil n/2\rceil}{\textstyle )$ 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.