Question

The English alphabets have 5 vowels and 21 consonants. How many words with two different vowels and two different consonants can be formed from the alphabet?

• A. \( 2100 \times 2! \)
• B. \( 210 \times 2! \)
• C. \( 210 \times 4! \)
• D. \( 2100 \times 4! \)

Hint:

Use combinations (\(\binom{n}{k}\)) for selection when order doesn't matter.
Use permutations (\(P(n,k)\) or \(n!\)) for arrangement when order matters.
Break the problem into steps: selection of vowels, selection of consonants, then arrangement of the selected letters.

Answer 1

The Correct Option is D

The task is to form words with 2 different vowels and 2 different consonants. Number of vowels = 5. Number of consonants = 21. Step 1: Select 2 different vowels from 5 vowels. Number of ways = \(\binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10\). Step 2: Select 2 different consonants from 21 consonants. Number of ways = \(\binom{21}{2} = \frac{21 \times 20}{2 \times 1} = 21 \times 10 = 210\). Step 3: Combine these selections. Number of ways to select the group of 4 letters (2 different vowels, 2 different consonants) = (Ways to select vowels) \(\times\) (Ways to select consonants) = \(10 \times 210 = 2100\). Step 4: Arrange these 4 selected distinct letters to form a word. The 4 selected letters are all different (2 different vowels, 2 different consonants, and vowels are different from consonants). Number of ways to arrange 4 distinct letters = \(4!\). \(4! = 4 \times 3 \times 2 \times 1 = 24\). Total number of words that can be formed = (Number of ways to select the group of 4 letters) \(\times\) (Number of ways to arrange these 4 letters) Total words = \(2100 \times 4!\). This matches option (d). \[ \boxed{2100 \times 4!} \]