Answer:
[tex]\textsf{There are $\boxed{14,400}$ 5-letter "words" with 2 vowels and 3 consonants.}[/tex]
[tex]\textsf{There are $\boxed{1,440}$ 5-letter "words" with a vowel at the end.}[/tex]
Step-by-step explanation:
The 10-letter word FORMULATED is made up of:
- 4 vowels: O, U, A, E
- 6 consonants: F, R, M, L, T, D
To determine the number of 5-letter "words", begin by considering the number of possible arrangements. This involves selecting the positions for the vowels. Once the positions for the vowels are chosen, those for the consonants are implicitly determined.
Out of a total of 5 positions, we need to choose 2 positions for the vowels. So, the total number of arrangements for 2 vowels and 3 consonants is calculated as follows:
[tex]\displaystyle \textsf{Number of arrangements}=\binom{5}{2}=\dfrac{5!}{2!(5-2)!}=\dfrac{5!}{2!\:3!}=10[/tex]
If the vowels and consonants cannot be repeated, there are:
- 4 choices for the first vowel and 3 choices for the second vowel.
- 6 choices for the first consonant, 5 choices for the second consonant, and 4 choices for the third consonant.
To find the total number of combinations of 5-letter "words" where each contains 2 vowels and 3 consonants, multiply the total number of possible arrangements by the number of ways to arrange each vowel and consonant:
[tex]10 \times 4 \times 3 \times 6 \times 5 \times 4 = 14400[/tex]
So, the total number of 5-letter words where each contains 2 vowels and 3 consonants is 14,400.
The number of arrangements where the two vowels are at each end of the “word" is one:
- Vowel, Consonant, Consonant, Consonant, Vowel
Again, we have 4 choices for the first vowel, 3 choices for the second vowel, 6 choices for the first consonant, 5 choices for the second consonant, and 4 choices for the third consonant. So the number of combinations where the two vowels are at each end of the “word" is:
[tex]1 \times 4 \times 3 \times 6 \times 5 \times 4 = 1440[/tex]
