# How many 12-letter strings are there that contain at least 3 vowels (a vowel is one of {a,e,i,o,u})?

Are these two answers equivalent?If not,why?

$$26^{12}-\left(\binom{21}{12}+\binom51\binom{12}1\binom{21}{11}+\binom51\binom{12}1\binom51\binom{11}1\binom{21}{10}\right)$$

$$\binom{24}{12}+\binom{25}{12}+\binom{26}{12}$$

-
Please check that I interpreted everything correctly. – Brian M. Scott Apr 9 '13 at 0:42
Those numbers aren’t even in the same ballpark: the first is much larger than the second. The first is over $9.5\times10^{16}$; the second is less than $1.76\times10^7$. – Brian M. Scott Apr 9 '13 at 0:44
Is there another way to solve this question? – Manuel Apr 9 '13 at 0:50
Are repeating letters (including vowels) allowed? For instance, does "xxxxxxxxxaaa" count? – MGA Apr 9 '13 at 0:54

Since there are $21$ consonants, there are $21^{12}$ $12$-letter strings with no vowels. To build a string with exactly one vowel, first pick a string of $11$ consonants, which you can do in $21^{11}$ ways; then pick one of the $12$ possible positions for the vowel, and pick which of the $5$ vowels to use. Altogether this results in $21^{12}\cdot5\cdot12$ strings with exactly one vowel. Finally, to build a string with exactly two vowels, pick a string of $10$ consonants, which you can do in $21^{10}$ ways; then pick $2$ of the $12$ positions for the vowels, which you can do in $\binom{12}2$ ways; and finally pick $2$ vowels, which you can do in $5^2$ ways. Altogether you get $21^{10}\binom{12}25^2$ strings with exactly two vowels.
Since there are $26^{12}$ strings altogether, your final answer is
$$26^{12}-21^{12}-60\cdot21^{11}-25\binom{12}221^{10}=26^{12}+39\cdot21^{11}-1650\cdot21^{10}=26^{12}-831\cdot21^{10}\;.$$