Assume that $e+f+g=n$.
Prove:
$P(n;e,f,g)=P(n-1;e-1,f,g) + P(n-1;e,f-1,g) + P(n-1;e,f,g-1)$
Where $P(n;e,f,g)$ is the tri-nomial-coefficient of n over e,f,g ($\frac{n!}{e!f!g!}$)
Combinatorially.
2026-03-25 23:51:02.1774482662
Trinomial equation - combinatorial explanation
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Hint: Imagine you have a line with $n$ blanks and an alphabet consisting of $\{E,F,G\}.$ Call $P(n;e,f,g)$ the number of ways that you create a word with exactly $e$ E's, $f$ F's and $g$ G's.
For the RHS assume that you are going to fix the last letter. You have $3$ options and in each option you will end with $n-1$ blanks.
For the LHS imagine first that every letter is different, in how many ways can you create a permutation of those $n$ letters? Now, you now that they are not different, you know that there are $e$ E's and so in how many ways can you just order those? Use multiplication principle.