Find $n$ when the coefficients of the $16^{th}$ and $26^{th}$ terms of $(1+x)^n$ are equal.

Question

Find $n$ when the coefficients of the $16^{th}$ and $26^{th}$ terms of $(1+x)^n$ are equal.

$16^{th}$ term coefficient: $^nC_{15}= \space ^nC_{n-15}$

$26^{th}$ term coefficient: $^nC_{25}= \space ^nC_{n-25}$

$\Rightarrow 15 = n-25 \Rightarrow n=40$ $\space$ or $\space$ $n-15=25 \Rightarrow n=40$

This is the correct answer. My question is since $^nC_{15}= \space ^nC_{25}$, why can't $n-15=n-25$? In this case $-15 \neq -25$, but if the algebra works would it be allowed? After all the question says the coefficients are equal. Thanks.

Answer 1

No, for the same reason that $15\neq 25$ even though $\binom{n}{15}=\binom{n}{25}$

The question tells you two terms that are in different positions, 16th and 26th. To find the correct answer of $40$, you use the symmetry property to change one of these, so that you have two representations of the same position: $15, n-25$ or $25, n-15$.

If you use the symmetry property to change both positions, you're back to talking about different positions, hence it doesn't work.