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Count arrangements with product, sum and restriction rules

Start a set of 8
Step 1 of 5

Which plan counts three positions filled from eight distinct cards without reuse?

Year 11, VCE Units 1 & 2, Chapter 8, 4 difficulty levels, No calculatorBuild a worksheet

Counting arrangements

First identify the positions and every restriction. Multiply successive choices in one arrangement, and add counts only for alternative cases that cannot occur together

\[N = c_1 \times c_2 \times \ldots \times c_k\]
  1. Multiply successive choices Use multiplication when all positions are filled as part of one arrangement \(\displaystyle 4 \times 3 \times 2 = 24\)
  2. Add disjoint cases Use addition when exactly one of several non-overlapping cases applies \(\displaystyle (3 \times 2) + (4 \times 2) = 14\)
  3. Reduce choices after use When repetition is forbidden, one fewer distinct object is available after each placement \(\displaystyle 6 \times 5 \times 4 = 120\)
  4. Count a required block Treat an adjacent group as one outer unit, then count its internal orders \(\displaystyle 4! \times 2! = 48\)

Worked example

Arrange 6 distinct labels in a row. A specified pair of labels must stay adjacent. How many arrangements are possible?

\[5! \times 2!\]
  1. \(\displaystyle 6 - 2 + 1 = 5\)The adjacent pair is one outer unit, so there are 5 outer units
  2. \(\displaystyle 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)Arrange the 5 outer units in a row
  3. \(\displaystyle 2! = 2 \times 1 = 2\)The two labels in the block can be ordered in 2 ways
  4. \(\displaystyle 5! \times 2! = 120 \times 2 = 240\)Multiply the outer arrangements by the internal block orders

Common mistakes

Adding successive choices
✕\(\displaystyle 8 + 7 + 6 = 21\)✓\(\displaystyle 8 \times 7 \times 6 = 336\)
Successive positions form one arrangement, so their choice counts must be multiplied rather than added
Reusing a placed object
✕\(\displaystyle 5 \times 5 \times 5 = 125\)✓\(\displaystyle 5 \times 4 \times 3 = 60\)
When repetition is forbidden, each placement leaves one fewer available object
Counting a circle as a line
✕\(\displaystyle 6! = 720\)✓\(\displaystyle (6 - 1)! = 5! = 120\)
For a circle with rotations treated as the same, fix one object and arrange the remaining objects

Difficulty levels

introShort slot products and unrestricted linear arrangements
standardRepetition rules, disjoint alternatives or one required adjacent block
challengeCircular, repeated-object or not-together restrictions with several cases
examChoose the appropriate counting model and calculate an exact whole-number count

Questions about this topic

When should choices be multiplied?+

Multiply when all choices are made in sequence as part of one arrangement. Each position is a further choice in the same outcome

When should counts be added?+

Add counts when the alternatives are separate cases and no arrangement belongs to more than one case

Why is a circular arrangement divided by rotations?+

Rotating every object around the circle does not create a new arrangement when rotations are treated as identical, so one object can be fixed as an anchor

Can I use a calculator?+

No. This topic is practised without a calculator

Which year level and chapter is this?+

Year 11, VCE Units 1 & 2, Mathematical Methods, chapter 8: Probability and counting

In this chapter

Finite events, probability assignments and complementsRepresent a multi-stage finite sample spaceCompound events, the addition rule and mutual exclusivityComplete and use a two-way probability tableConditional probability from exact representationsDistribution of a finite random variableDesign and interpret a probability simulationCount restricted selections and calculate exact probabilitiesCount arrangements with product, sum and restriction rulesTest exact independenceExact probability of repeated independent successesMulti-stage probability, multiplication and total probabilityProbability from a uniform area modelProbabilities of sequences of independent events