Find the value of n if the coefficients of the sixth and sixteenth terms in the expansion of (a+b)ⁿ are equal.

The coefficient of the $(r+1)$-th term in $(a+b)^n$ is $\binom{n}{r}$.
For the sixth term, $r=5$, so the coefficient is $\binom{n}{5}$.
For the sixteenth term, $r=15$, so the coefficient is $\binom{n}{15}$.

Given $\binom{n}{5} = \binom{n}{15}$.
Using the property $\binom{n}{r} = \binom{n}{k} \implies r=k$ or $r+k=n$.
Since $5 \neq 15$, we must have $n = 5+15$.

The value of $n$ is $\boxed{20}$.

: Expand the following: $(a+b-c)^3$.

Let $X = (a+b)$. Then the expression is $(X-c)^3$.

The expansion is $\boxed{a^3 + b^3 - c^3 + 3a^2b + 3ab^2 - 3a^2c - 3b^2c + 3ac^2 + 3bc^2 - 6abc}$.

**: Write down in its simplest form the complete