Question

In how many ways can you create 5-letter words using the letters of the word 'AMBIDEXTROUS' such that there is no repetition of letters and there are exactly 3 consonants and 2 vowels?

Hint:

When forming words from a set of letters, first choose the required number of consonants and vowels, then multiply by the number of ways to arrange the selected letters.

Answer 1

Step 1: Identifying consonants and vowels in the word 'AMBIDEXTROUS'.
The word 'AMBIDEXTROUS' consists of the following letters:
A, M, B, I, D, E, X, T, R, O, U, S.
- Vowels: A, I, E, O, U (5 vowels)
- Consonants: M, B, D, X, T, R, S (7 consonants)
Step 2: Choosing the consonants and vowels.
We need to choose 3 consonants and 2 vowels from the available consonants and vowels:
- Number of ways to select 3 consonants from 7 consonants: \( \binom{7}{3} \)
- Number of ways to select 2 vowels from 5 vowels: \( \binom{5}{2} \)
\[ \binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \] \[ \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10 \]
Step 3: Arranging the selected letters.
After selecting the 3 consonants and 2 vowels, we have 5 letters in total. These 5 letters can be arranged in \( 5! \) ways: \[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]
Step 4: Calculating the total number of arrangements.
The total number of ways to form the 5-letter word is the product of the ways to select the consonants, the ways to select the vowels, and the number of ways to arrange the letters: \[ \text{Total Ways} = \binom{7}{3} \times \binom{5}{2} \times 5! = 35 \times 10 \times 120 = 42,000 \]
Step 5: Conclusion.
Thus, the total number of ways to create 5-letter words from 'AMBIDEXTROUS' with exactly 3 consonants and 2 vowels is 42,000. The correct answer is 42,000.