How many 8-letter words contain exactly 5 vowels (a,e,i,o,u)? What if repeated letters were not allowed?
This question has two parts to be answered.
The first part is,"How many 8-letter words contain exactly $5$ vowels (a,e,i,o,u)?" so what I did did $8$ boxes, it can have all $5$ vowels and I am left with $3$ boxes left. So the answer is $(8C5)\cdot 5^5 \cdot 21^3$
now for the last part of the question,"What if repeated letters were not allowed?" so again I used the 8 boxes, and I can't use the same letters are vowels, so it's 5! because I can't repeat the letters. Each vowel is 5 boxes of the 8; $(8C5)$. Now I have 3 boxes left, this is where I am stuck, the answer has $(8C5)\cdot 5! \cdot (21P3)$. Why are they using permutation for the last three boxes?
Is it correctly done thus far?
Best Answer
It says no letters are repeated (which means none of the $8$ letters).
Once we find all combinations of $5$ places for vowels in the word, which is $\displaystyle {8 \choose 5}$, we only need to now permute vowels and consonants within themselves. Hence the answer.
Another way to look at it is that we choose $3$ consonants in $\displaystyle {21 \choose 3}$ ways and then permute all $8$ letters of the word, which is $8!$
So the answer is $\displaystyle {21 \choose 3} \cdot 8! \ $ which is same as the answer in your book.