<s>[INST] Show that $p \in \left[\frac{4^m}{2\sqrt{m}},\frac{4^m}{\sqrt{2m+1}}\right]$ 
If the number of ways in which $m$ identical apples can be put in $2m$ boxes, so that no box contains more than one apple, is $p$, prove that $$p \in \left[\frac{4^m}{2\sqrt{m}},\frac{4^m}{\sqrt{2m+1}}\right]$$   

I did this as follows :  
Let the number of apples in $i^{th}$ box be $x_{i}$, then,  
$$\sum_{i=1}^{2m}x_{i}=m$$  
where $x_{i} \in \{0,1\}$ and the number of ways would be the number of solutions to this equation, which is equal to the coefficient of $x^m$ in $(1+x)^{2m}$ i.e, $p=\dbinom{2m}{m}$  
However, I can't prove that $p \in \left[\frac{4^m}{2\sqrt{m}},\frac{4^m}{\sqrt{2m+1}}\right]$. Can we prove this inequality without induction? Also, is my method correct ?   
Any help will be appreciated.
Thanks
 [/INST] Here is another rather elementary answer.

Let $a_m=\frac{1}{4m}\binom{2m}{m}$. We show the following is valid
\begin{align*}
\frac{1}{2\sqrt{m}}\leq  a_m \leq \frac{1}{\sqrt{2m+1}}\qquad\qquad m\geq 1\tag{1}
\end{align*}

We start similarly to @CuriousGuest
\begin{align*}
a_m&=\frac{1}{4^m}\binom{2m}{m}=\frac{(2m)!}{4^mm!m!}=\frac{(2m)!!(2m-1)!!}{4^mm!m!}\\
&=\frac{(2m-1)!!}{2^mm!}=\frac{(2m-1)!!}{(2m)!!}\\
&=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdots\frac{2m-1}{2m}
\end{align*}

Now we consider $a_m^2$ and obtain
  \begin{align*}
a_m^2&=\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\frac{5}{6}\cdots\frac{2n-1}{2n}\cdot\frac{2n-1}{2n}\\
&>\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdots\frac{2m-2}{2m-1}\cdot\frac{2m-1}{2m}\\
&=\frac{1}{4m}
\end{align*}
  Thus
  \begin{align*}
a_m>\frac{1}{2\sqrt{m}}
\end{align*}
which proves the left inequality of (1)

$$$$

In order to show the right inequality we observe
  \begin{align*}
a_m&=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdots \frac{2m-1}{2m}\\
&<\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdots \frac{2m}{2m+1}\\
&=\frac{2}{1}\cdot\frac{4}{3}\cdot\frac{6}{5}\cdots \frac{2m}{2m-1}\,\cdot\,\frac{1}{2m+1}\\
&=\frac{1}{a_m}\cdot\frac{1}{2m+1}
\end{align*}
  We conclude
  \begin{align*}
a_m^2&<\frac{1}{2m+1}\\
a_m&<\frac{1}{\sqrt{2m+1}}
\end{align*}
  which proves the right inequality of (1).

</s>