Trying to understand the formula for counting multiset permutations 
In how many ways can we plant $5$ red, $3$ yellow and $2$ white flowers in a row?

The answer is $\frac {10!}{(5! \cdot 3! \cdot 2!)}$. So it looks like we are dividing out the redundant permutations. Does that presuppose that $5! \cdot 3! \cdot 2!$ are factors of $10!$ ? If all the flowers were the same color, then there'd be $10!$ ways to plant them in a row and no subsets of the flowers would permute within themselves. That is $5! \cdot 3! \cdot 2!$ would not be a factor of $10!$. So without subset permutations we get $10!$ ways to plant the flowers in a row, shouldn't we get more ways when the subsets permute? 
 A: Consider choosing $10$ positions for flowers.
First consider red ones, you want to choose $5$ positions for red ones out of $10$ positions, the order doesn't matter since flower with same color are identical. Therefore there are $10 \choose 5$ way.
Then consider yellow ones, similarly you choose $3$ positions out of the rest of $5$ positions, so number of ways is $5 \choose 3$.
Finally white ones have the only choice to fill in the blanks.
The total number of ways is $${10 \choose 5} \times {5 \choose 3} = \frac{10!}{5! \cdot 3! \cdot 2!}$$
You can switch the order, check out whether it yields the same result.
A: If all the flowers where the same there would be $1$ way to put them in a row.
If all the flowers where different there would be $10!$ ways.
In this case there are $\frac{10!}{5!\cdot3!\cdot2!}$ because the $10!$ permutations can be seperated into groups of $5!\cdot3!\cdot2!$ permutations each that all look the same (Given a permutation you can permute the red flowers internally in $5!$ ways, these permutations leave the appearance of the arrangement unchanged, in other words if we consider each of the $10!$ arrangements that would be possible if the flowers where different we are counting each of the arrangements $5!\cdot 3!\cdot 2!$ times, when we consider many of the flowers are identical.
