How many unique permutations exist for the practice word ALOMOMOLA?

Conditions

• Word length is 9.
• Letter frequencies: $A=2$, $L=2$, $M=2$, $O=3$.

Reasoning, step by step

1. Count the total letters: 9.
2. Calculate $9! = 362,880$.
3. Identify repeat groups: $A (2)$, $L (2)$, $M (2)$, $O (3)$.
4. Set up the denominator: $2! \cdot 2! \cdot 2! \cdot 3! = 2 \cdot 2 \cdot 2 \cdot 6 = 48$.
5. Divide the total factorial by the denominator: $362,880 / 48 = 7,560$.

Example

The card 'Practice: ALOMOMOLA' states: 'The nine letters contain two each of A, L and M, and three O’s... there are 7,560 distinct arrangements.' Formula: $\frac{9!}{2!\,2!\,2!\,3!}=7560$.

Common misconceptions

• Missing one of the pairs of repeated letters in the denominator.
• Calculating $3!$ as 3 instead of 6.

Watch the explanation

Connected concepts

Explore next

• What if the word had four 'O's instead of three?
• How does the complexity increase with more unique letter types?

Related questions