Question 37 of 42intermediate🔧 ApplyNumerical3 marks

The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet?

Correct Answer

Step 1: Number of ways to select 2 different vowels from 5 vowels = 5C2=5!2! 3!=10^5C_2 = \frac{5!}{2!\,3!} = 10.

Step 2: Number of ways to select 2 different consonants from 21 consonants = 21C2=21!2! 19!=21×202=210^{21}C_2 = \frac{21!}{2!\,19!} = \frac{21 \times 20}{2} = 210.

Step 3: Number of combinations of 2 vowels and 2 consonants = 10×210=210010 \times 210 = 2100.

Step 4: Each combination has 4 distinct letters, which can be arranged in 4!=244! = 24 ways to form different words.

Step 5: By the multiplication principle, total number of words = 2100×24=504002100 \times 24 = 50400.

Hence, the required number of words is 50400.

Exercise: Miscellaneous Exercise on Chapter 6 | Q: 6 | (Page 24)
For More Understanding

Explanation

This question follows the same pattern as Example 20 in the textbook. First, we select the required vowels and consonants using combinations because order does not matter at the selection stage. Then we arrange the selected letters using permutations because the order of letters in a word matters. The multiplication principle combines the two steps to give the final count.

Solution Steps

  1. Step 1: Select 2 vowels from 5: ^5C_2 = 10.

  2. Step 2: Select 2 consonants from 21: ^21C_2 = 210.

  3. Step 3: Multiply selections: 10×21010 \times 210 = 2100 combinations.

  4. Step 4: Arrange each combination of 4 distinct letters: 4! = 24.

  5. Step 5: Total words = 2100×242100 \times 24 = 50400.

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