Arrangements with Repeated Objects

Not examined in 0606. The Cambridge IGCSE Additional Mathematics 0606 (2025–2027) permutations content excludes arrangements with repetition of identical objects. This page is here for completeness and for students who also sit A Level (9709) or SPM, where the rule does appear. It is not a 0606 exam skill, so leave it to last if you are only preparing for 0606.

When some of the objects being arranged are identical, the ordinary $n!$ overcounts, because swapping two identical letters produces an arrangement you can’t tell apart. The fix is to divide by the factorial of each group of repeats.

The rule

The number of distinct arrangements of $n$ objects, where one object repeats $p$ times, another $q$ times, and so on, is: $\frac{n!}{p!\,q!\,\cdots}$

Each repeated group’s internal orderings ($p!$ of them) are indistinguishable, so dividing removes the duplicates.

A worked example

How many distinct arrangements are there of the letters of the word $\text{BANANA}$? There are $6$ letters in total, with $\text{A}$ repeated $3$ times and $\text{N}$ repeated $2$ times. $\frac{6!}{3!\,2!} = \frac{720}{6 \times 2} = \frac{720}{12} = 60$ So there are $60$ distinct arrangements.

Without dividing, $6! = 720$ would count each genuinely different arrangement $3! \times 2! = 12$ times over.

Why the division works

Imagine the three A’s were labelled $\text{A}_1, \text{A}_2, \text{A}_3$. Then $6!$ would be right. But the labels aren’t real, and the $3! = 6$ ways of ordering the three A’s all look the same, so we’ve counted each real arrangement $6$ times. Dividing by $3!$ corrects it; similarly $2!$ for the two N’s.

Common mistakes

• Forgetting to divide at all (giving $n!$ and overcounting).
• Dividing by the wrong factorial, use the count of each repeated letter, not the number of distinct letters.
• Adding the factorials in the denominator instead of multiplying them.

Full topic context: Permutations & Combinations notes, and Arrangements & Selections.