Arrangements of a Word with Repeated Letters (Beyond 0606)

Not examined in 0606. Cambridge IGCSE Additional Mathematics 0606 (2025–2027) excludes permutations with repetition of identical objects. This page is extension practice, useful for A Level (9709) and SPM, not a 0606 exam question. For the P&C content that is examined, see the topic pillar.

If every letter were distinct, a word of $n$ letters would have $n!$ arrangements. But swapping two identical letters gives the same word, so you divide by the factorial of each repeat count to remove the overcounting.

Find the number of different arrangements of all the letters of the word $\text{BANANA}$.

The working

Step 1, count the letters and the repeats. $\text{BANANA}$ has $6$ letters:

• $\text{A}$ appears $3$ times,
• $\text{N}$ appears $2$ times,
• $\text{B}$ appears $1$ time.

Step 2, apply $\dfrac{n!}{p!\,q!\cdots}$: $\frac{6!}{3!\,2!}$

Step 3, evaluate: $= \frac{720}{6 \times 2} = \frac{720}{12} = 60$

There are $60$ different arrangements.

Where the marks are won and lost

• Divide by the factorial of every repeated letter: here both $3!$ (for the three A’s) and $2!$ (for the two N’s).
• The letter appearing once contributes $1! = 1$, so it makes no difference, but count the total letters correctly ($6$, giving $6!$ on top).
• Simplify carefully: $6! = 720$, $3!\,2! = 12$.

Common mistakes

• Dividing by only one of the repeats (e.g. $\frac{6!}{3!}$).
• Using $5!$ on top by miscounting the letters.
• Multiplying the factorials on the bottom incorrectly ($3! \times 2! = 12$, not $6$).

Topic home: Permutations & Combinations pillar. More: Worked examples.