CUET UG 2025 — General Test Quantitative Reasoning

The word OFFICE contains 6 letters: O, F, F, I, C, E

Vowels: O, I, E (3 vowels, all different)

Consonants: F, F, C (3 consonants, F repeats twice)

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To find arrangements where vowels never come together:

Vowels never together = Total arrangements - Vowels always together

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The word has 6 letters with F repeating twice.

Total arrangements $= \dfrac{6!}{2!}$=2!6!​

$$= \dfrac{720}{2}$$=2720​

$$= 360$$=360

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Treat all vowels (O, I, E) as one single block [OIE].

The units to arrange are: [OIE], F, F, C (4 units with F repeating twice)

Arrangements of these 4 units $= \dfrac{4!}{2!}$=2!4!​

$$= \dfrac{24}{2}$$=224​

$$= 12$$=12

Arrangements of vowels within the block $= 3!$=3!

$$= 6$$=6

Total arrangements with vowels together $= 12 \times 6$=12×6

$$= 72$$=72

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Arrangements where vowels never come together $= 360 - 72$=360−72

$$= 288$$=288

Therefore, the answer is 288.