Given that the coefficients of the $6^{th}$ term and $16^{th}$ term in the Expansion of $(x + y) ^n$ are equal. Find the value of $n$.

Question

The Expansion gives me: $6^{th}$ term = $\displaystyle\binom{n}{5} x^{n-5}y^5$ and $16^{th}$ term = $\displaystyle\binom{n}{15} x^{n-15}y^{15}$. What do I do next?

Answer 1

$\dfrac{n!}{5!(n-5)!}=\dfrac{n!}{15!(n-15)!} $. So $n$ is $20$.

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For completeness $\left(\text{Courtesy:} \: \textbf{Did} ,\: \textbf{gandalf61} ,\: \textbf{bof} ,\: \textbf{Francis}\right)$ $\\\text{(*I agree this explanation still has a lot of details missing)}$

The coefficient of $x^ky^{n−k}$ is equal to the coefficient of $x^{n−k}y^k$. So we have $n−5=15n−5=15$ and hence $ n=20$.

$\underline{\text{Other solutions to}\:\: \displaystyle\binom{n}{5}=\binom{n}{15}\quad(**)}$

The binomial coefficient is defined as $$\displaystyle \binom{n}{k}=\begin{cases}\dfrac{n!}{k!(n-k)!} &0\leq k \leq n \\ 0 &\text{otherwise} \end{cases}$$ So by definition, $n=0,1,2,3,4$ are also solutions to $(**)$. $$\displaystyle \binom{n}{5}=\binom{n}{15}\implies \displaystyle \dfrac{15!}{5!}=\dfrac{(n-5)!}{(n-15)!}$$ If $z\in\mathbb{C}$ and $a,b\in\mathbb{Z}\:$ we have $$\dfrac{\Gamma(z-a+1)}{ \Gamma(z-b+1)}=(-1)^{b-a}\frac{\Gamma(b-z)}{ \Gamma(a-z)}$$ Here $a=5, b=15$. $$\dfrac{\Gamma(z-5+1)}{ \Gamma(z-15+1)}=(-1)^{15-5}\frac{\Gamma(15-z)}{ \Gamma(5-z)}=\prod_{i=5}^{14}(i-z)$$ So $\displaystyle \dfrac{15!}{5!}= \prod_{i=5}^{14}(i-z)$ has $10$ roots by fundamental theorem of algebra which are complex (requires more details).

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So for our question here it is reasonable to assume $n$ to be positive, $n\geq k$ and the expansion of $(x+y)^n$ to have $n+1$ terms.

So $\fbox{$n$=$20$}$