CUET UG 2025 — General Test Quantitative Reasoning

The word is DELETE: D-E-L-E-T-E

Total letters = 6 letters

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Count each letter:

• D appears 1 time
• E appears 3 times
• L appears 1 time
• T appears 1 time

The letter E repeats 3 times.

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When letters repeat, the formula for arrangements is:

$$\text{Number of arrangements} = \frac{n!}{n_1! \times n_2! \times ...}$$Number of arrangements=n1​!×n2​!×...n!​

Where $n$n is the total number of letters and $n_1, n_2...$n1​,n2​... are the number of times each repeating letter appears.

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For the word DELETE:

$n = 6$n=6 (total letters)

E repeats 3 times

$$\text{Arrangements} = \frac{6!}{3!}$$Arrangements=3!6!​

$$= \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1}$$=3×2×16×5×4×3×2×1​

$$= \frac{720}{6}$$=6720​

$$= 120$$=120

Therefore, the letters of DELETE can be arranged in 120 different ways.