How many 8-letter words with no repeated letters can be constructed using the 26 letters of the alphabet if each word contains 3, 4, or 5 vowels?
There are 26 letters in the English language, including 5 vowels: \(\{a, e, i, o, u\}\).
Let’s split up the cases:
- Words with 3 vowels. We can choose 3 vowels in \(\binom{5}{3} = 10\) ways. The remaining 5 letters can be chosen from the remaining 21 letters (only consonants) in \(\binom{21}{5} = 20349\) ways. That gives us a total of \(\binom{5}{3} \cdot \binom{21}{5} = 203490\) ways to pick the 8 letters. Those 8 letters can be permuted in \(8! = 40320\) ways for the total of \(\binom{5}{3} \cdot \binom{21}{5} \cdot 8! = 8204716800\) ways.
- Words with 4 vowels. We can choose 4 vowels in \(\binom{5}{4} = 5\) ways. The remaining 4 letters can be chosen from the remaining 21 letters (only consonants) in \(\binom{21}{4} = 5985\) ways. That gives us a total of \(\binom{5}{4} \cdot \binom{21}{4} = 29925\) ways to pick the 8 letters. Those 8 letters can be permuted in \(8! = 40320\) ways for the total of \(\binom{5}{4} \cdot \binom{21}{4} \cdot 8! = 1206576000\) ways.
- Words with 5 vowels. We can choose 5 vowels in \(\binom{5}{5} = 1\) ways. The remaining 3 letters can be chosen from the remaining 21 letters (only consonants) in \(\binom{21}{3} = 1330\) ways. That gives us a total of \(\binom{5}{5} \cdot \binom{21}{3} = 1330\) ways to pick the 8 letters. Those 8 letters can be permuted in \(8! = 40320\) ways for the total of \(\binom{5}{5} \cdot \binom{21}{3} \cdot 8! = 53625600\) ways.
The grand total is the sum of the partial sums:
\[\binom{5}{3} \cdot \binom{21}{5} \cdot 8! + \binom{5}{4} \cdot \binom{21}{4} \cdot 8! + \binom{5}{5} \cdot \binom{21}{3} \cdot 8! = 9464918400\]