(B) The string "$KCET$ $2025$" contains $8$ characters: $K, C, E, T, \text{space}, 2, 0, 2, 5$. However, assuming the space is ignored and we consider

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(B) The string "$KCET$ $2025$" contains $8$ characters: $K, C, E, T, \text{space}, 2, 0, 2, 5$. However, assuming the space is ignored and we consider the $8$ characters: $K, C, E, T, 2, 0, 2, 5$.Fix $K$ at the first position and $5$ at the last position.The remaining $6$ characters are $C, E, T, 2, 0, 2$.Among these $6$ characters, the digit $2$ repeats twice.The number of ways to arrange these $6$ characters is $\frac{6!}{2!} = \frac{720}{2} = 360$ ways.

Class 11 Mathematics Permutation and Combination Definition of permutation, Number of permutations with or without repetition, Conditional permutations

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How many ways can you arrange all the characters in "$KCET$ $2025$" such that the arrangement starts with $K$ and ends with $5$?

KCET 2026 All KCET

A
$720$
B
$360$
C
$120$
D
$180$

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Class 11

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Permutation and Combination

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Definition of permutation, Number of permutations with or without repetition, Conditional permutations

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How many ways can you arrange all the characters in "$KCET$ $2025$" such that the arrangement starts with $K$ and ends with $5$?

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