How many 9 letter strings are there that contain at least 3 vowels? I'm studying for my exams and stuck on this one question.
The way I'm thinking of doing this is by:
$$26^9 - \binom{26}3-\binom{26}2-\binom{26}1-\binom{26}0= 5,429,503,676,728$$
But that seems very, very wrong. Any help would be appreciated. Thanks!
 A: One way to see that it’s almost certainly wrong is to notice that it doesn’t in any way take into account the fact that there are $5$ vowels. On the other hand, the idea of starting with all $26^9$ strings and throwing out the ones with fewer than $3$ vowels is good. Specifically, you want to throw out the ones that have $0,1$, or $2$ vowels. (You went a step too far.)
There are $5$ vowels and $21$ consonants, so there are $21^9$ strings composed entirely of consonents $-$ i.e., with $0$ vowels. 
To make a string with exactly $2$ vowels, you must choose which $2$ of the $9$ positions are to be filled with vowels, then choose vowels for those $2$ positions, and finally choose consonants for the other $7$ positions. You can do that in $\binom92\cdot5^2\cdot21^7$ ways.
How many ways are there to make a string with exactly one vowel?
A: The strategy you used  is basically right. A number of the details are not. There are $26^9$ strings. We subtract the number of "bad" ones.
There are $21^9$ all consonant ($0$ vowel) strings.
Now count the $1$ vowel strings. The location of the vowel can be picked in $\binom{9}{1}$ ways. That location can be filled with a vowel in $5^1$ ways, And the remaining $8$ locations can be filled with consonants in $21^8$ ways, for a total of $\binom{9}{1}5^121^8$.
Now we count the strings with  exactly $2$ vowels. Where the vowels will go can be chosen in $\binom{9}{2}$ ways. These locations can be filled with vowels in $5^2$ ways. And for each of thsse ways, the remaining $7$ spots can be filled with consonants in $21^7$ ways.
