Question

There are 5 different caps $c_1,c_2,c_3,c_{4}and{c}_5$ and 5 different boxes $B_1,B_2,B_3,B_{4}and{B}_5$. The capacity of each box is sufficient to accommodate all the 5 caps. If $B_1$ and $B_5$ have the caps $C_1$ and $C_5$ among themselves, in how many ways can you arrange the caps among the 5 boxes?

• A. 400
• B. 485
• C. 500
• D. 365

Answer 1

The correct option is C 500

$B_1$ and $B_5$ can be filled up in 2 $\times$ 2 = 4 ways and $B_2,B_3,B_4$ can be filled up in $=5\times 5\times 5=125ways$.
Hence the required number of ways $=4\times 125=500$