Trouble understanding concept: number of strings of 6 characters with exactly two vowels. So in order to count the number of strings with exactly two vowels, the equation would be $5^2 * C(6,2) * 21^4$. I understand that we use the $C(6,2)$ to count the total combination of positions that can be made for two spots in a string of 6 characters. However, I'm not sure if I understand why we do not impose the same limitation for the number of consonants, $21^4$. For the remaining 4 spots, why am I not required to find the number of positions that could be made for the 4 spots in a string of 6 characters, or $C(6,4)$? Wouldn't we be overcounting when we count the permutations of positions for the consonants? Or is it because $C(6,4)$ is the symmetry of $C(6,2)$, and therefore results in a matching combination for the consonants?
 A: After you choose $2$ places for vowel, you have $4$ more places to choose from.  Hence it is actually
$$5^2 \cdot C(6,2) \cdot 21^4 \cdot C(4,4)= 5^2 \cdot C(6,2) \cdot 21^4$$
In another word, the moment you choose the positions for vowels, you know that you have to fill the rest for consonants.
A: To avoid complication let's say you want to use lower case letters in your string.

When you choose $2$ positions to be vowels you are, in fact, saying that the remaining $4$ positions are not vowels. If you then intend to fill these remaining $4$ spaces with letters then they must be consonants. 
e.g The choice of vowel positions here actually implies the consonant/vowel ($C/V$) "configuration" beneath it: 
$$\begin{array}{cccccc}\_&\_&V&\_&\_&V\\\downarrow&\downarrow&\downarrow&\downarrow&\downarrow&\downarrow\\C&C&V&C&C&V \end{array}$$
There are 
$$C(6,2)=\binom{6}{2}=15$$ 
different configurations of $4$ consonants and $2$ vowels similar to the one shown here. 
For each different configuration we can replace each of the $2$ $V$s with vowels $a,e,i,o,u$ in $5$ ways, so that's $5^2$ ways for vowels. 
Then for each of those we can replace each of the $4$ $C$s with consonants $b,c,d,\ldots ,z$ in $21$ ways, so that's $21^4$ ways for consonants.
In total, for each configuration of $C$s and $V$s we can replace $C$s and $V$s in $5^2\cdot 21^4$ ways. 
Since we have $\binom{6}{2}=15$ different $C/V$ configurations and $5^2\cdot 21^4$ ways to replace $C$s and $V$s in each different configuration there are
$$\binom{6}{2}\cdot 5^2\cdot 21^4$$
total strings with $2$ vowels and $4$ consonants.
