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Im trying to understand an example of multinomial theorem, and have a question.

Lets say I want to expand $(a+b+c)^2$.

I apply the theorem that says $(a+b+c)^2= \sum_{i=1}^{2} \frac{2!}{n_1!n_2!n_3!} \cdot a^{n_1} \cdot b^{n_2} \cdot c^{n_3}$ where $n_1+n_2+n_3=2$.

According to the solution I want to find all combinations of $n_1+n_2+n_3$ that satisfies they sum up to 2, and the solution only give these for combinations.

$n_0=0, n_1=0, n_2=2$

$n_0=0, n_1=1, n_2=1$

$n_0=1, n_1=0, n_2=1$

$n_0=1, n_1=1, n_2=0$

However, should there also be these combinations aswell?

$n_0=2, n_1=0, n_2=0$

$n_0=0, n_1=2, n_2=0$

uoiu
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1 Answers1

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What you've written is correct. I'm not sure if you've misread the solution or if there is a typo in your source, but all six combinations that you identify are valid for the equation in the OP, and they correspond to the six terms of $(a+b+c)^2$.

You can verify that this is correct without using the theorem, by writing $(a+b+c)^2 = ((a+b)+c)^2$ and then performing FOIL twice, first with $(a+b)$ and $c$. (Note F = $(a+b)^2$, hence the need for a second one).

Eric Nathan Stucky
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