# Given that $\binom{16}r =\binom{16}{2r + 1}$, what is the value of $r$?

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

From what I understand:

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

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

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

• algebra.com/algebra/homework/Permutations/… Dec 15, 2018 at 9:24
• Use ${n\choose m}={n\choose n-m}$ and $r\ne 2r+1$ to find $r=16-(2r+1)\Rightarrow r=5$. Dec 15, 2018 at 9:32
• Don't forget that any negative integer $r$ is a trivial solution. Dec 16, 2018 at 8:46
• @user10354138, and any integer greater than 16. Dec 17, 2018 at 14:50

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 (and also in @AmbretteOrrisey's answer), 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.

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 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 $$16-r=2r+1$$$$\therefore$$$$3r=15$$$$\therefore$$$$r=5 .$$