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Quantitative Aptitude

If B 3 can ha...

Question

If $B_{3}$ can have only $C_{1}$ or $C_{5}$ , in how many ways can you arrange the caps such that all boxes have one cap?

A

480

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B

420

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C

48

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D

88

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Solution

The correct option is C 48
$B_{3}$ can be filled in 2! ways and rest can be filled in 4! ways.
Hence, required number of ways $= 2 \times 4! = 48.$

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