Counting

Counting without double-counting

Define exactly what is being counted. Check overlap and whether order matters before reaching for a formula.

Make the cases exhaustive and control overlap

Adding two counts works directly only when the two groups are disjoint. If an object appears in both groups, subtract the overlap once. For several choices, decide whether changing the order creates a new outcome.

  1. Describe one outcome precisely: an integer, an ordered tuple, a selection or an arrangement.
  2. Choose disjoint cases, or explicitly correct overlapping ones.
  3. For a small version of the problem, list outcomes and compare with the method before using it at full size.

Worked example

Multiples of either number

How many integers from 1 to 60 inclusive are divisible by 3 or 5, including integers divisible by both?

There are 60/3 = 20 multiples of 3 and 60/5 = 12 multiples of 5. An integer divisible by both is a multiple of 15; there are 60/15 = 4 of these.

The sum 20 + 12 counts those four twice. Subtract them once:

20 + 12 − 4 = 28.

If the question instead asked for integers divisible by exactly one of 3 and 5, the overlap would need to disappear altogether. That count would be 20 + 12 − 2 × 4 = 24. Read the wording before choosing the correction.

Your turn

A restriction on one variable

How many ordered triples (x, y, z) of non-negative integers satisfy x + y + z = 7 and x ≤ 3?

Show a hint

Take x = 0, 1, 2 and 3 separately. For a fixed x, how many choices for y are possible, and what then determines z?

Show the full solution

For a fixed x, y can be any integer from 0 to 7 − x, giving 8 − x choices. Each determines exactly one valid z.

The four disjoint cases give 8 + 7 + 6 + 5 = 26 ordered triples.

Another route uses stars and bars. Without the restriction, the count is C(9, 2) = 36. Invalid triples have x ≥ 4. Setting u = x − 4 gives u + y + z = 3, with C(5, 2) = 10 solutions. Thus 36 − 10 = 26. Here C(n, 2) = n(n − 1)/2.

The first method makes the restriction visible and needs no memorised formula. Both methods count ordered triples, so swapping unequal values between positions changes the outcome.

Check the four cases by listing their possible y values. Explain why none of the cases overlap.

What to practise next

Start with sets and overlapping events, then move to restricted counting. These are general problem-solving exercises; check your test’s current specification for its assessed content.

The examples and explanations on this page are part of Arij Asad’s teaching guides. Linked papers and worksheets belong to their named creators and publishers. PMT’s topic collections include exam-board questions and other credited materials.