Question:medium

The number of 4 letter words (with or without meaning) that can be formed from the eleven letters of the word 'EXAMINATION' is

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Always check frequency of repeated letters. Break into cases: all distinct, one pair, two pairs, triple, quadruple.
Updated On: Apr 22, 2026
  • 2454
  • 3025
  • 3462
  • 4096
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
This is a permutation and combination problem involving repeated letters. The word 'EXAMINATION' has 11 letters: A, A, I, I, N, N, E, X, M, O, T.
There are 8 distinct letters: {A, I, N, E, X, M, O, T}.
There are 3 pairs of identical letters: {AA, II, NN}.
Step 2: Key Formula or Approach:
We must consider three cases for selecting 4 letters:
Case 1: 2 pairs of identical letters.
Case 2: 1 pair of identical letters and 2 distinct letters.
Case 3: 4 distinct letters.
Step 3: Detailed Explanation:
Case 1: 2 pairs (e.g., AAII)
Ways to choose 2 pairs from 3 available: \( ^3C_2 = 3 \).
Arrangements: \( \frac{4!}{2!2!} = 6 \).
Total for Case 1: \( 3 \times 6 = 18 \).

Case 2: 1 pair and 2 different (e.g., AAIN)
Ways to choose 1 pair from 3 available: \( ^3C_1 = 3 \).
Ways to choose 2 distinct letters from the remaining 7 distinct letters: \( ^7C_2 = 21 \).
Arrangements: \( \frac{4!}{2!} = 12 \).
Total for Case 2: \( 3 \times 21 \times 12 = 756 \).

Case 3: 4 distinct letters (e.g., EXAM)
Ways to choose 4 letters from 8 distinct ones: \( ^8C_4 = 70 \).
Arrangements: \( 4! = 24 \).
Total for Case 3: \( 70 \times 24 = 1680 \).

Grand Total:
Summing the cases: \( 18 + 756 + 1680 = 2454 \).
Step 4: Final Answer:
The total number of 4-letter words is 2454.
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