Arrangements with a Restriction: Letters Together

“Must be together” questions use the block method: glue the joined items into one unit, arrange everything, then multiply by the ways the block can be ordered internally. It converts a restriction into an ordinary arrangement.

How many different arrangements are there of the six letters of the word $\text{NUMBER}$ in which the two vowels ($\text{U}$ and $\text{E}$) are next to each other? [4]

The working

All six letters of NUMBER are distinct, which keeps the counting clean.

Step 1, glue the vowels into a block. Treat $[\text{UE}]$ as one unit. Now arrange $5$ units: the block plus $\text{N}, \text{M}, \text{B}, \text{R}$: $5! = 120 \quad \text{(M1, M1)}$

Step 2, arrange the vowels inside the block. $\text{U}$ and $\text{E}$ can be $\text{UE}$ or $\text{EU}$: $2! = 2 \quad \text{(M1)}$

Step 3, multiply: $5! \times 2! = 120 \times 2 = 240 \quad \text{(A1)}$

Where the marks are won and lost

• Arranging $5$ units, not $6$: gluing the vowels reduces the count of things to arrange by one. Using $6!$ ignores the restriction.
• The internal $2!$ is essential, the block $[\text{UE}]$ and $[\text{EU}]$ are different arrangements. Forgetting it gives $120$, exactly half the correct answer.
• All letters distinct means no division for repeats. (If a letter repeated, you would divide by the factorial of its count.)

Common mistakes

• Using $6!$ and forgetting the letters are joined.
• Omitting the $\times 2!$ for the block’s internal order.
• Over-counting by treating the vowels as always in a fixed order.

Full method: Arrangements & Selections notes. See also Factorials. Topic home: Permutations & Combinations pillar.