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
• B. 420
• C. 48
• D. 88

Answer 1

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 $=2\times 4!=48.$