KCET · Maths · Permutation Combination
How many ways can you arrange all the letters and numbers in "KCET 2025" which start with K and end with \(5\)?
- A \(720\)
- B \(360\)
- C \(120\)
- D \(180\)
Answer & Solution
Correct Answer
(B) \(360\)
Step-by-step Solution
Detailed explanation
The given alphanumeric string is KCET 2025.
The letters and numbers present are K, C, E, T, \(2\), \(0\), \(2\), \(5\).
The total number of characters to be arranged is \(8\).
The first position is fixed for K and the last position is fixed for \(5\).
The remaining \(6\) positions are to be filled by the remaining \(6\) characters: C, E, T, \(2\), \(0\), \(2\).
Among these \(6\) characters, the digit \(2\) is repeated twice.
The number of ways to arrange these \(6\) characters is \(\dfrac{6!}{2!}\).
\(\dfrac{6!}{2!} = \dfrac{720}{2} = 360\)
Answer: \(360\)
The letters and numbers present are K, C, E, T, \(2\), \(0\), \(2\), \(5\).
The total number of characters to be arranged is \(8\).
The first position is fixed for K and the last position is fixed for \(5\).
The remaining \(6\) positions are to be filled by the remaining \(6\) characters: C, E, T, \(2\), \(0\), \(2\).
Among these \(6\) characters, the digit \(2\) is repeated twice.
The number of ways to arrange these \(6\) characters is \(\dfrac{6!}{2!}\).
\(\dfrac{6!}{2!} = \dfrac{720}{2} = 360\)
Answer: \(360\)
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