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): $\dfrac{8!}{3!}$
A permutation is an arrangement of multiple objects in a particular order taken a few or all at a time. The formula for permutation is as follows:
\(^nP_r = \frac{n!}{(n-r)!}\)
nPr = permutation
n = total number of objects
r = number of objects selected