Finding the coefficient How to find the coefficient of $a^3b^4c^5$ in the expansion of $(ab+bc+ca)^6$
 A: You need to find how many ways of multiplicating 6 terms of the forms $ab$, $bc$ and $ca$ will give you $a^3b^4c^5$. Let's describe such a way:
Since the exponent of $c$ is $5$, and you're multiplicating 6 terms, exactly one of them must not have a $c$, so you must have exactly one $ab$ in the multiplication. Now, the exponent of $b$ is $4$, so you'll be needing $3$ other therms that have a $b$ and are of the forms $bc$ or $ca$. Then there must be 3 terms $bc$ in the multiplication, and the other 2 are $ca$'s.
What we have prooved by now is that if you're choosing elements of the forms $ab$, $bc$ and $ca$ such that their product is $a^3b^4c^5$, then you must have chosen exactly 1 $ab$, 3 $bc$'s and 2 $ca$'s.
Now imagine what would happen if you used distributivity to expand $(ab +bc +ca)^6=(ab+bc+ca)\cdots(ab+bc+ca)$, maintaining the order in which you multiply. You'd have many terms that are actually equal to $a^3b^4c^5$ but are only written in a different form (such as $(ab)(bc)(bc)(bc)(ca)(ca)$ and $(bc)(ab)(bc)(bc)(ca)(ca)$). You only need to count how many of them are there. I think it's pretty clear that each of these different therms will only appear once after you've done the expansion.
You have $6!$ ways of ordenating the "distinct" terms $ab$, $bc$, $bc$, $bc$, $ca$, $ca$, $ca$. Now you have to remove the "duplicates", so you'll divide by $3!$ and then by $2!$, since there are 3 terms $bc$ and 2 terms $ca$. That wields a total of $6!/(3!2!)=60$ ways of writing $a^3b^4c^5$ as a product of $ab$, $bc$ and $ca$.
Therefore, the coefficient of $a^3b^4c^5$ in the expansion of $(ab+bc+ca)^6$ is $60$.
PS.: I never really understood permutations and combinations. I find it very natural to imagine you're "removing duplicate therms".
A: We don't have too many possibilities. Consider the $a$'s, exactly $3$ of them is needed. All cannot come from $ab$ because then $c^2$ would be left. Similarly, all can't come from only $ca$'s. Similarly, if $ab$ is chosen from $2$ places, it's $a^3b^2c$, and $b^2c^4$ is left for the $bc$'s.
Assume we have $1$ component where $ab$ was chosen, then it's already $(ab)(ca)^2=a^3bc^2$. So, $(bc)^3$ is left, that can be done in $6\cdot\displaystyle\frac{5\cdot 4}2=60$ ways.
Another way is, that consider we have $x$ pieces of $ab$'s, $y$ of $bc$'s and $z$ of $ca$'s, then we have the following equations to solve:
$$\begin{align}  x+z &= 3 \\
x+y&=4\\ y+z&=5
\end{align}$$
