Combinations
The total number of ways of selecting 4 letters from the word 'EXAMINATION' is
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About This Question
- Subject
- mathematics
- Chapter
- permutations and combinations
- Topic
- combinations
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
136
- Catalog Letters: E(1), X(1), A(2), M(1), I(2), N(2), T(1), O(1).
- Distinct types: 8 (E, X, A, M, I, N, T, O).
- Doublet pairs: 3 (AA, II, NN).
- Case 1: All 4 different: Select 4 types from 8.
- C(8,4)=248×7×6×5=70.
- Case 2: 2 alike, 2 different:
- Choose 1 pair from 3 types: C(3,1)=3.
- Choose remaining 2 types from remaining 7: C(7,2)=21.
- Total Case 2 = 3×21=63.
- Case 3: 2 alike of one kind, 2 alike of another:
- Choose 2 pairs from the 3 available types.
- C(3,2)=3.
- Grand Total: 70+63+3=136.
This hard difficulty mathematics question is from the chapter permutations and combinations, covering the topic of combinations. It appeared in the 2025 exam.
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