Answer:
- combinations: 1200
- vowel at each end: 120
Explanation:
You want to know the number of 5-letter combinations of 3 consonants and 2 vowels can be formed from the letters of "FORMULATED", and the number that have a vowel at each end.
Combinations
Since the problem statement uses the word "combinations" instead of "permutations", we take it to mean that "FORMU" is to be considered the same as "FOMRU" and "FUMRO", which have the same letters.
Since the position of the vowels seems to matter, either of the above is considered different from "FORUM" where the vowels are in different places.
The number of combinations of 3 consonants from the 6 in "FORMULATED" is 6C3 = 6!/(3!(6 -3)!) = 20, and the number of combinations of 2 vowels of the 4 given is 4C2 = 4!/(2!(4-2)!) = 6.
The possible arrangements of 2 vowels and 3 consonants in a group of 5 letters is 5C2 = 5!/(2!(5-2)!) = 10.
So, the combinations of 3 consonants and 2 vowels in with vowels in the different possible positions is ...
20·6·10 = 1200
There can be 1200 different 5-letter combinations.
Vowel position
There is only one of the 10 possible arrangements of consonants and vowels that has the vowels at each end. The number of such arrangements is 1/10 of the total, or (1/10)(1200) = 120.
120 of the letter combinations will have a vowel at each end.
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Additional comment
More often, we're interested in the number of "words", where letter order matters. If that is intended to be the case, then the number of 5-letter "words" is 6P3·4P2·5C2 = 14400, and the number that have vowels at each end is 1440.
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