Question

The number of 4-letter words formed from the English alphabet such that there are exactly 2 vowels and 2 consonants and no vowel is repeated, but consonants may be repeated is:

• A. \(^4C_2\times ^5C_2\times (21)^2\)
• B. \(^4C_2\times ^5C_2\times ^{21}C_2\)
• C. \(^4C_2\times ^5P_2\times ^{21}P_2\)
• D. \(^4C_2\times ^5P_2\times (21)^2\)

Hint:

Choose positions for vowels, arrange distinct vowels, then fill consonant places with repetition allowed.

Answer 1

The Correct Option is D

Step 1: Pattern approach
Pick which two positions hold vowels: there are $6$ ways, which is $^4C_2$.

Step 2: Fill each type
First vowel place: $5$ choices. Second vowel place: $4$ choices (no repeat), so $5\times 4 = ^5P_2$. Each consonant place: $21$ choices, so $21\times 21$.

Step 3: Combine
By the multiplication principle the number of words is $^4C_2\times{}^5P_2\times 21^2$, which is option (D).

Final Answer:
Option D. \[ \boxed{\text{(D)}\ {}^4C_2\times {}^5P_2\times (21)^2} \]