Question:

1 red, 3 white, 2 blue flags in a line, no adjacent same colour, ends different. How many arrangements?

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Break arrangement problems by fixing constraints (like end colours) and count valid permutations.
Updated On: Aug 5, 2025
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The Correct Option is B

Solution and Explanation

Brute force counting possible sequences satisfying: no adjacent same colour, ends different colour. After arranging whites in separate spots and interleaving red/blue, only 4 sequences possible. \[ \boxed{4} \]
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