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Question

There are 5 different caps c1,c2,c3,c4 and c5 and 5 different boxes B1,B2,B3,B4 and B5. The capacity of each box is sufficient to accommodate all the 5 caps. If B3 can have only C1 or C5, 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
B3 can be filled in 2! ways and rest can be filled in 4! ways.
Hence, required number of ways =2×4!=48.

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