I have proved that $\sin A \sin B =\sin x \sin y $.
How to show $\sin x \sin y \leq \sin^2 (\frac C2)$ ?

From AM-GM,
I have got: $\sin x \sin y \leq \sin^2 (\frac C2) \cos^2 (\frac {x-y}2) $

Can someone ELI5 the reasoning behind this: There is a class of 30 students and 4 class committee positions (pres, vice pres, secretary, treasurer). I'm trying to find every combo of students and committee members. The only stipulation is that any student can serve any role, but they can't serve more than one role at once.

I thought the number of combinations would have been 30 choose 4, but apparently, it is 30! (factorial) I'm scratching my head as to why.

I tried approaching this without Stirling's in mind, we can show that the limit exists so suppose that it equals $L$. Now, let's find $L$.
$\begin{align*}
\frac {a_n}{b_n}&=\frac {\sum_{k=1}^n k!-(k-1)!}{n!}+\sum_{k=1}^n \frac{(k-1)!}{n!}\\
&=\frac {n!-0!}{n!}+\frac{(n-1)!}{n!}.\frac {a_{n-1}}{b_{n-1}}\to 1+0=1
\end{align*}$

@copper: The solution given in the book is along the following lines:Given $\epsilon>0$, there is a $\delta, 0<\delta<1$, such that if $0<|x|<\delta$, then
$|f(1-x[1/x])|<\epsilon$. Taking $n$ so large as $1/n<\delta$. For $0<s<\frac 1{(n+1)}$, set $x=\frac {1-s}n$, then $\frac 1{n+1}<x<\frac 1n$. Thus $[1/x]=n$.
If $0<s<\frac 1{n+1}$, then $|f(s)|<\epsilon$. For $s<0$, one can proceed analogously.

I don't understand my mistake here https://math.stackexchange.com/questions/4363927/proof-verification-given-lim-x-to-0-fracf2x-fxx-0-and-lim-x-to/4363947?noredirect=1#comment9115580_4363947
Please let me know what I did wrong here.

For any $\epsilon\gt0$, there is an $x_\epsilon$ so that for $x\le x_\epsilon$, we have $\left|\frac{f(2x)-f(x)}{x}\right|\le\epsilon$. For any $x\le x_\epsilon$,
$$
\begin{align}
\left|\frac{f(x)-\lim\limits_{y\to0}f(y)}x\right|
&=\left|\sum_{k=0}^\infty\frac{f(2^{-k}x)-f(2^{-k-1}x)}{2^{-k-1}x}2^{-k-1}\right|\\
&\le\sum_{k=0}^\infty\epsilon2^{-k-1}\\[9pt]
&=\epsilon
\end{align}
$$