Given that the alphabet soup contains the vowels (A, E, I, O, U) with each of the vowels appearing five times.
This means that the total number of letters in the bowel soup is 25 letters.
The number of five-letter words that can be formed from the bowel of 25 letters assuming there are no restrictions is given by
[tex]^{25}P_5= \frac{25!}{(25-5)!} \\ \\ = \frac{25!}{20!} =25\times24\times23\times22\times21 \\ \\ =6,375,600[/tex]