<s>[INST] Binomial theorem and multinomial coefficient I have a question to which I could not find an answer in the forum.
I am trying to solve a bracelet problem. Actually I already solved it but with some kind of a cheating (using wolframaplha to expand the Dihedral group  formula). The problem that I try to solve is: how many different bracelets we can create from 12 beads (3 red, 4 green 5 blue).
The Dihedral group is:
$$ D_n = \frac{f_1^{12} + f_2^6 + 2f_3^4 + 2f_4^3 + 2f_6^2 + 4f_{12}^1 + 6f_1^2f_2^5 + 6f_2^6}{24} $$
and if we replace in the formula for the three colors we got:
$$ D_n = \frac{(x + y + z)^{12} + (x^2 + y^2 + z^2)^6 + 2(x^3 + y^3 + z^3)^4 + 2(x^4 + y^4+z^4)^3 + 2(x^6 + y^6 + z^6)^2 + 4(x^{12} + y^{12} + z^{12}) + 6(x + y + z)^2(x^2 + y^2 + z^2)^5 + 6(x^2 + y^2 + z^2)^6}{24}$$
From here by using wolframalpha to expand each part I found that only:
$(x + y + z)^{12}$ 
and 
$6(x + y + z)^2(x^2 + y^2 + z^2)^5$
can make $x^3y^4z^5$
From there I also get the coeficent before them and found the answer 1170 bracelets from 3 red, 4 green and 5 blue beads.
However, I have two questions because I do not want to cheat with wolframalpha:
1. How I can found which parts of the formula could expand to $x^3y^4z^5$
2. How I can use multinomial coefficient for calculating them without expanding. I know that by using it I do not need to expand the formula. I need for example the formula for something hard:
$$(x^2 + y^2 + z^2)^5$$
The problem that I have is solved, I just need an explanation how can I approach the hard part without using a tool.
Thanks for the answers.
 [/INST] For the first question, note that for example in $(x^2+y^2+z^2)^6$, each term will be of the form $kx^{2a}y^{2b}z^{2c}$ for some integers $k,a,b,c$.  So it's not possible to have an exponent of $3$ or $5$. A similar consideration eliminates every term but the two you have mentioned.
For multinomial coefficients in the example you give, $$\left(x^2+y^2+z^2\right)^5=\sum_{a+b+c=5}\binom{5}{a,b,c}x^{2a}y^{2b}z^{2c},\tag1$$ where the sum is over all ordered triples $(a,b,c)$ of nonnegative integers such that $a+b+c=5$.
In general, we have the multinomial formula: $$\left(x_1+x_2+\cdots+x_k\right)^n=\sum_{n_1+\cdots+n_k=n}\binom{n}{n_1,\dots,n_k}x_1^{n_1}\cdots x_k^{n_k}\tag2$$ 
If you have an expression with powers of the variables, just substitute them in the formula.  That's how to get $(1)$ from $(2)$.  
EDIT
So to finish the necklace problem, from the first term we get $$\frac1{24}\binom{12}{3,4,5}=1155$$.  Consider the second term, $$6(x + y + z)^2\left(x^2 + y^2 + z^2\right)^5$$  Since the second factor will give only even powers, we need odd powers of $x$ and $z$ from the first term.  Since the exponent is $2$, these powers can only be $1$.  From the second term then we need $x^{2a}y^{2b}z^{2c}=x^2y^4z^4$ so $a=1,b=2,c=2$.  That gives $$\frac6{24}\binom{2}{1,0,1}\binom{5}{1,2,2}=15$$ 
</s>