Cfrgtkky

Solve
$$displaystylebinom{16}r =binom{16}{2r + 1}.$$

From what I understand:

$$displaystylefrac{16!}{r!(16-r)!}=frac{16!}{(2r+1)!(16-(2r+1))!}$$

$$implies r!(16-r)! = (2r+1)!(16-(2r+1))!$$

But I am stuck at this point.
If there is any straightforward way to solving this problem, please describe it.

This is a great question! If all you have learnt is the definition of
$binom nm$ as
$$
binom nm = frac{n!}{m!(n-m)!}tag{$dagger$}
$$
then this problem seems quite imposing. How is one to check for which values of $r$
the equation
$$
r!(16-r)! = (2r+1)! (16-(2r+1))!
$$
holds?!

However, if you have the correct tool in hand then this problems opens up to your efforts very easily. The basic idea is given by @farruhota in the comments, but I believe there is one key ingredient still missing, so I will elaborate on that in this answer.

Now, it is easy to see from the definition of $binom nm$ that $$binom nm = binom{n}{n-m}.$$
Indeed, just plug into the formula $(dagger)$
and check that LHS = RHS.

However, you cannot immediately use this to solve
$$r = 2r+1$$
and thereby get the desired value of $r$. Because, perhaps there also exist some other integers $0 leq r < l leq n$ such that $binom nr = binom nl$ is true? One still needs to rule out that possibility in order to be able to argue as @farruhota does in the comments.

As it happens, it is not too difficult to prove that $binom nr = binom nl$ holds if and only if $r = l$ or $r = n - l$. So, here goes: say $0 leq r < l leq n$ and
$$
binom nr = binom nl.
$$
Then, using the definition we get
$$
begin{align}
&&frac{n!}{r! (n-r)!} &= frac{n!}{l!(n-l)!}\
iff&& l!(n-l)! &= r!(n-r)!\
iff&& frac{l!}{r!} &= frac{(n-r)!}{(n-l)!}\
iff&& l(l-1)(l-2) cdots (r+1) &= (n-r)(n-r-1)(n-r-2) cdots (n-l+1).tag{$*$}
end{align}
$$

Now, in the last equation, both the LHS and the RHS are a product of $l-r$ consecutive integers. When can they be equal to each other? Well, for any positive integers $a,b,k$, we have
$$
a(a+1)(a+2) cdots (a+k) = b(b+1)(b+2) cdots (b+k) iff a = b. qquad (text{Why? Check this!})
$$

And, et voila! We can now conclude that $(*)$ holds if and only if
$$
r+1 = n-l+1 iff r = n-l
$$
as we wanted to show.

With this result in hand, it is now easy to solve this problem. I believe you can now take it from here. :)

Binomial coefficients are duplicated at the 'mirror-image' argument - by which I mean $$binom{n}{k}=binom{n}{n-k} .$$ The binomial 'function' of order n, with k considered to be a variable, is exactly symmetrical about the centreline. So in this case we have $$16-r=2r+1$$$$therefore$$$$3r=15$$$$therefore$$$$r=5 .$$