Use the multinomial theorem to expand $(x+y+z)^4$.
To calculate the number of terms, you apply the following formula: $\binom{n+r-1}{n}$. Here $n=4$ and $r=3$. So $\binom{6}{4}=15$. I don't understand how they are getting $15$ terms.
The Multinomial Theorem states:
$(x_1+\cdots+x_r)^n = \sum\limits_{n_1+...+n_r = n}\binom{n}{n_1,...,n_r}\cdot x_1^{n_1} \cdots x_r^{n_r}$.
The sum runs over all possible partitions $n_1,...,n_r$ such that $n_1+...+n_r=n$
There are only $3$ sets of numbers to get $n=4$:
1. $(4,0,0) \implies$ Here there are $3$ objects but $2$ are the same. So there are $3!/2! = 3$ permutations
2. $(3,1,0) \implies 3! = 6$ Permutations
3. $(2,2,0) \implies 3!/2! = 3$
$3+6+3 = 12$.
So for example, for item one you would have: $\binom{4}{4,0,0} x^4 y^0 z^0 + \binom{4}{0,4,0} x^0 y^4 z^0 + \binom{4}{0,0,4} x^0 y^0 z^4$
For item $2\\$:
$(3,0,1), (3,1,0), \\
(1,3,0), (0,3,1), \\
(1,0,3), (0,1,3)$
For item $3\\$: $(2,2,0), (0,2,2,), (2,0,2)$
What are the other $3$ terms? Thank you!
