Lonely Coefficients

Let $p+q+r=n$ , then the coefficent of $a^pb^qc^r$ in the expansion of $(a+b+c)^n$ is equal to $\dfrac{n!}{p!q!r!}$ , so it is also the same as the coefficent of $a^qb^qc^r$ .

So, if there exists a lonely coefficent, that means $p=q=r$ , and we cannot "trade the places of the exponential".

We have $n$ must be a multiple of $3$ (because $n=p+q+r=3p$ ), but we are not sure about if every multiple of $3$ will work.

Fortunely, the maximum value of $\dfrac{n!}{p!q!r!}$ occurs when $p=q=r$ , so of course that coefficent is lonely for every $n$ is a multiple of $3$

Hence, the probability equals $\dfrac13$

The coefficients of an expanded trinomial correspond to the numbers in different layers of Pascal's pyramid , in which each layer is a triangle with three-way symmetry that can be obtained by adding the three numbers above it. The first few layers are:

The only time a "lonely coefficient" appears is in the center of a triangular layer, which is only possible if the triangular number $\frac{(n + 1)(n + 2)}{2}$ (for each Layer $n$ ) is not divisible by $3$ . Inspecting the equation, we see that this can only occur if $n \equiv 0 \pmod{3}$ (as $n + 2$ is divisible by $3$ for $n \equiv 1 \pmod{3}$ and $n + 1$ is divisible by $3$ for $n \equiv 2 \pmod{3}$ ).

Therefore, ${\displaystyle \lim_{n\to\infty}} \frac{A(n)}{n} = \frac{1}{3}$ , so $p = 1$ and $q = 3$ , and $p + q = 1 + 3 = \boxed{4}$ .