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: Count the letters available
The English alphabet has \(5\) vowels and \(21\) consonants.

Step 2: Choose positions
We need exactly two of the four places for vowels. This can be done in \(^4C_2\) ways. The other two places are for consonants.

Step 3: Fill the vowel places
Vowels cannot repeat, so the two vowel places are filled in \(^5P_2\) ways (choose and arrange).

Step 4: Fill the consonant places
Consonants may repeat, so each of the two places has \(21\) choices, giving \(21^2\).

Step 5: Multiply
\[ \text{Total} = {}^4C_2 \times {}^5P_2 \times (21)^2 \]

Step 6: Why other options fail
Options (A) and (B) use \(^5C_2\), which does not arrange the vowels. Options (B) and (C) use \(^{21}C_2\) or \(^{21}P_2\), which forbid repeated consonants, but the question allows them.

Final Answer:
The count is 4C2 x 5P2 x 21^2. \[ \boxed{\text{(D)}\ {}^4C_2\times {}^5P_2\times (21)^2} \]