Count the letters in $\text{LETTER}$

The word $\text{LETTER}$ has 6 letters: $L, E, T, T, E, R$

Here,

• Vowels: $E, E$ (2 vowels, identical)
• Consonants: $L, T, T, R$ (4 consonants, where $T$ repeats twice)

Arrange the consonants first

We arrange $L, T, T, R$.

Since the two $T$'s are identical, the number of distinct arrangements is $\frac{4!}{2!}=\frac{24}{2}=12$

Create gaps for placing the vowels

For any arrangement of the 4 consonants, there are 5 possible gaps where vowels can be placed so that no two vowels are adjacent: $\_\; C\; \_\; C\; \_\; C\; \_\; C\; \_$

So the number of gaps is 5.

Place the two vowels so that they never come together

Since the two vowels are both $E$, they are identical. To ensure they never come together, we must place them in two distinct gaps out of the 5 gaps.

Number of ways: $\binom{5}{2}=10$

Total number of valid words

Multiply the number of consonant arrangements by the number of ways to choose gaps for the vowels: $12 \times 10 = 120$

Compare with stored correct answer

Derived answer = $120$

Stored correct answer = $120$

Hence, they agree.