There are five different boxes with building materials. In

Hmm, I didn't get any of the answer, but here is how I did it..
Say if boxes are A,B,C,D,E
If A is on fifth floor, you have 5 floors left with four boxes. So for this case, you get P(5,4) = 2*3*4*5 = 120
If you have five boxes, you should get 120*5 = 600.
Maybe someone else can try.

this was my line of thought as well. I'm waiting for someone to come in and show us how it's done...

I am also getting 600 as the answer.

Lets say we have five boxes A, B, C, D, E.

And 6 floors F1, F2, F3, F4, F5, F6

Either of 5 boxes can be kept on F5 floor hence 5 ways.
Let's keep F6 as empty 1st.
For each of the above 5 way, we have, 4 boxes placed in on floors F1, F2, F3 and F4 in 4! ways = 4.3.2 = 24.
Hence we have 24.5 = 120 ways of keeping boxes on 5 floors when F6 is empty.
Now we can repeat this exercise for each floor except F5. So we have 120 ways for each 5 floors which comes as 120 * 5 = 600 ways. I am sure we are not missing anything. Hence I think the answer is 600.

Vlad77, Could you please check if you have not done any typo ..

What is the OA ?

there is no way you can have over 3,000 possible permutations in this situation.

but at the same time that's odd they wouldn't have what we think is the correct answer even listed.

it is 3125, and C (3.125 I think is a typo). 5x5x5x5x5