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Problem-solving strategy

Count the opposite

‘At least one’ is usually messy: it might be one, two, three or more. Its opposite, ‘none’, is a single clean case. Count or find the probability of the opposite, then subtract from the total.

The same idea works with areas (shaded region = big shape minus pieces) and with conditions (good arrangements = all arrangements minus bad ones).

When to try it

  • The words ‘at least’, ‘not all’, ‘not next to each other’ or ‘contains’ appear.
  • The cases you want overlap in complicated ways, but the cases you do not want are simple.
  • A shaded area is irregular but the shapes around it are standard.

Watch out: Make sure you know the total exactly (and that its outcomes are equally likely, for probability). Subtracting from the wrong total is the commonest slip.

Two worked examples

Try each one first. The hints and the full solution are underneath.

Problem J06

Junior · Number theoryShort answer

Mia writes the numbers 1 to 30. She circles every number that is a multiple of 2 or a multiple of 3, but not a multiple of 5. How many numbers does she circle?

Hint

First count the multiples of 2 or 3, then remove the ones that are multiples of 5.

Second hint

Multiples of 2 or 3 up to 30: 15 + 10 − 5 = 20. Which of these are multiples of 5?

Full worked solution

Answer: 16

  1. Multiples of 2 from 1 to 30: 30 ÷ 2 = 15.
  2. Multiples of 3: 30 ÷ 3 = 10.
  3. Multiples of both (that is, of 6) were counted twice: 30 ÷ 6 = 5.
  4. Multiples of 2 or 3: 15 + 10 − 5 = 20.
  5. Remove those that are also multiples of 5: 10, 15, 20, 30 (4 numbers; 5 and 25 are not multiples of 2 or 3).
  6. Circled numbers: 20 − 4 = 16.

Why this works: This is inclusion–exclusion: when two lists overlap, add them and subtract the overlap once, so nothing is counted twice.

Where it leads: Combining ‘or’ and ‘but not’ conditions is inclusion–exclusion; a Venn diagram with three circles keeps track of every region.

Strategy: Count the opposite, Organised cases

Problem I08

Intermediate · CombinatoricsMultiple choice

In how many ways can the six letters of the word LEVELS be arranged so that the two Es are not next to each other?

Hint

Count all arrangements (remember the repeated letters), then subtract those with EE together.

Second hint

All arrangements: 6!/(2! 2!) = 180. With EE glued: 5!/2! = 60.

Full worked solution

Answer: C, 120

  1. LEVELS has six letters: L twice, E twice, V once, S once.
  2. All arrangements: 6! ÷ (2! × 2!) = 720 ÷ 4 = 180 (divide out swaps of identical letters).
  3. Arrangements with the two Es together: glue them into one block EE, leaving five items L, L, V, S, EE.
  4. Those arrange in 5! ÷ 2! = 60 ways.
  5. Not together: 180 − 60 = 120 (C).

Why this works: ‘Not together’ = all − together, and ‘together’ is counted by gluing. Dividing by factorials of repeated letters removes arrangements that look identical.

Where it leads: ‘Not together’ is easiest by complement; the ‘gaps method’ (place the other letters, then drop the Es into gaps) is an alternative.

Strategy: Count the opposite

Practise: 58 problems that use count the opposite

Other strategies

Organised cases · Working backwards · Invariants · Extremal principle · Pigeonhole principle · Parity and remainders · Symmetry · Spot the pattern and generalise · Proof techniques

All strategy guides · Extension & competition maths