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You can use Lagrange inversion to explicitly solve

$$x^5-x-a=0\qquad (*)$$

(yes, a fifth degree equation, gasp). More precisely, it yields an infintite infinite series expansion

$$x=-\sum_{k\geq 0}\binom{5k}{k}\frac{a^{4k+1}}{4k+1}$$

for the root of $(*)$ which is $0$ at $a=0.$ Although this isn't combinatorics, I'd gladly devote a class in any subject I teach to be able to derive it, because by Bring–Jerrard, any quintic equation can be reduced to this form, and you get a solution of something that many people believe, albeit for differing reasons, to be unsolvable.

$$x^5-x-a=0\qquad (*)$$
$$x=-\sum_{k\geq 0}\binom{5k}{k}\frac{a^{4k+1}}{4k+1}$$
for the root of $(*)$ which is $0$ at $a=0.$ Although this isn't combinatorics, I'd gladly devote a class in any subject I teach to be able to derive it, because by Bring–Jerrard, any quintic equation can be reduced to this form, and you get a solution of something that many people believe, albeit for differing reasons, to be unsolvable.