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

Solution

Maths sir · 4 steps

Sort the letters into pairs and singles first, then count each pattern of 4 letters separately.

1. List the letters. EXAMINATION has A, I and N twice each, and E, X, M, T, O once each. That makes 8 different letters, and only 3 of them can appear twice in a word.
2. Split into cases. No letter can appear three times, so a 4-letter word is one of three types: all different, one letter repeated with two others, or two letters repeated. Choose the letters first, then arrange them.

   Pattern	Choose the letters	Arrange them	Words
   All 4 different	C(8, 4) = 70	4! = 24	70 × 24 = 1680
   One pair + 2 different	pair: 3 ways, others: C(7, 2) = 21	4!/2! = 12	3 × 21 × 12 = 756
   Two pairs	C(3, 2) = 3	4!/(2! 2!) = 6	3 × 6 = 18

3. Why 7, not 5. With one pair fixed, the other two letters can be any of the remaining 7 different letters, including the other repeatable ones. Each of those is used only once in this case.
4. So. Total = 1680 + 756 + 18 = 2454.

The common slip: 7920. Treating all 11 letters as different gives 11 × 10 × 9 × 8 = 7920. That counts swaps of the two A's, two I's or two N's as new words.

The idea to keep. When a word has repeated letters, sort the letters into types first. Count each case as (ways to choose the letters) × (arrangements, divided by 2! for each repeated pair), then add the cases.

Answer from the NTA final answer key. Solution by PrepWiser.