How many rational elements are in the multinomial? $(\sqrt[3]{x}+\sqrt[5]{x}+2\sqrt{x})^5$
According to the multinomial theorem, $\sum_{i+j+k=5}^{}\frac{5!}{i!j!k!}2^k\ x^{\frac{1}{3}i+\frac{1}{5}j+\frac{1}{2}k}$
So I think it's rational when $k \ \in \ \mathbb{N}$ and $(\frac{1}{3}i+\frac{1}{5}j+\frac{1}{2}k)\ \in \ \mathbb{N}$ and $i+j+k=5$
Which satisfy only when $i=3,j=0,k=2$ and $j=5,i=k=0$ so if's correct then are only 2 rational elements in the sum iteration?
Additional question
Needing to find how many elements are in the multinomial, so what I figured is finding in how many for the equations $a + b + c= 5$ to be formed? which is $D(3,5) = C(3+5-1,5)$ but I am not sure about it
I agree with the analysis in the first part of your query. There is one ambiguity:
In the case of $i=3,k=2$, this can be regarded as one element or the sum of $\binom{5}{3} = 10$ elements, depending on your interpretation of the word element.
Depending on your interpretation of the word element, the second part of your query can be regarded as a Stars and Bars problem, where you want the number of non-negative integer solutions to
$$i + j + k = 5.$$
The enumeration is
$$\binom{5 + [3-1]}{[3-1]} = \binom{7}{2} = 21.$$
This agrees with the number that you (the OP) computed.