Ideally, what (s)he meant by smiley faces were when $b < a \leq c | b < c \leq a$, and frowny faces were if $b > a \geq c | b > c \geq a$.
Neutral face is when $a = b = c$. Smirks are all other possibilities of $a, b, c.$
Number of neutral faces: $n$ ways to choose $a$, and since $a=b=c,$ there are $n$ ways.
Number of smirks: $\binom{3}{2} \cdot n(n-1)$ because there are $n$ ways to choose $a,$ $n-1$ ways to choose $b$ or $c,$ and you could have any one of these be first so you multiply $n(n-1)$ by $\binom{3}{2}$ or $\binom{3}{1},$ either is same. So you have $3(n)(n-1)$ ways.
Number of smiley faces: $P\binom{n}{3} = 3 \cdot \binom{n}{3}.$
Number of frowny faces: Symmetric to number of smiley faces = $3 \cdot \binom{n}{3}.$
Number of ways to choose $a, b, c:$ $n \cdot n \cdot n = n^3.$
Add these up and you get $n^3 = n + 3n(n - 1) + 6 \binom{n}{3}$
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