6!×3!
6!×3
We must choose 3 positions out of 8 to place the vowels (in fixed order), and fill the rest with the 5 consonants.
Number of ways to choose 3 vowel positions from 8: $^8C_3$
Number of ways to arrange the remaining 5 consonants: $5!$
Total number of valid arrangements: $^8C_3 \times 5! = \dfrac{8!}{3!}$
Correct answer is (D): $\dfrac{8!}{3!}$
Given:
Word is VERTICAL, which has 8 letters.
Vowels in the word: E, I, A (3 vowels)
Step 1: Choose positions for the 3 vowels out of 8 letters: $^8C_3$
Step 2: Arrange the remaining 5 consonants in those 5 positions: $5!$
Total number of ways: $^8C_3 \times 5! = \dfrac{8!}{3!}$
Correct option is (D): $\dfrac{8!}{3!}$
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