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0606/22 Oct/Nov 2025, Question 5: Permutations: Counting Arrangements

Cambridge IGCSE Additional Mathematics 0606 · Paper 22 (Calculator) · Calculator · 5 marks

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The question, in short

From 5 letters, 4 numbers and 3 symbols, form 6-character passwords with no repeats; count passwords under three different sets of conditions.

Summarised in our own words. Use the official paper for the exact wording and any diagrams.

Part (a) · 1 mark

Count all possible 6-character passwords using distinct characters chosen from the 12 available.

  1. 1There are 12 different characters
  2. 2¹²P₆ = 12 × 11 × 10 × 9 × 8 × 7 = 665 280

Answer665 280

Part (b) · 2 marks

Count 6-character passwords with distinct characters where the first and last characters are both symbols.

  1. 1First and last characters: 3 × 2 ways to choose symbols
  2. 2Middle 4 characters from the remaining 10: 10 × 9 × 8 × 7
  3. 33 × 2 × 10 × 9 × 8 × 7 = 30 240

Answer30 240

Part (c) · 2 marks

Count passwords starting with a symbol then number or number then symbol, and ending with two letters.

  1. 1First two characters: symbol then number (3 × 4) or number then symbol (4 × 3): 2 × 3 × 4 = 24
  2. 2Last two characters are letters: 5 × 4 = 20
  3. 3Middle two characters from the 12 − 4 = 8 remaining: 8 × 7 = 56
  4. 424 × 56 × 20 = 26 880

Answer26 880

Stuck on a step? Read the free lesson: Permutations: Counting Arrangements

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Also searched as: 0606/22/O/N/25 · 0606_w25_qp_22 · 0606 w25 22 · question 5.
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