How many arrangements of length $12$ formed by different letters (no repetition) chosen from the $26$-letter alphabet are there that contain the five vowels $(a,e,i,o,u)$?

I know that there are $12$ spaces and $5$ vowels must be placed somewhere in those twelve spots with other letters in the other $7$ places. After the $5$ vowels are placed there are $(21*7) + (20*6) + (19*5) + (18*4) + (17*3) + (16*2) + (15*1)$ combinations for the remaining $7$ places. Is this correct? How do I calculate the arrangements of the $5$ vowels?


You can select $7$ of the $26-5=21$ non-vowels (consonants and semi-vowels), and then you can permute the resulting $12$ letters in $12!$ ways, so there are

$$ \binom{21}712!=55\,698\,306\,048\,000 $$



Pick the five spaces the vowels go into $(_{12}C_5)$ then pick the order from left to right ($5!$).

Then pick the seven consonants $(_{21}C_7)$ and then choose the order from left to right in the word $(7!)$.

The total number of words is the product of these four terms.


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