The problem can be solved in relatively simpler way without using combinations formula but by simply using counting method.
We have to fill 3 places by selecting 3 students from 4 groups of 4 such that no student is from same group.
Lets create groups:
1st group - a,b,c,d
2nd group - e,f,g,h
3rd group - i,j,k,l
4th group - m,n,o,p
The first position can be filled in 16 ways (a,b....p), 2nd can be filled in 12 ways - since 2nd student should not be from the same group (e,f....p), 3rd position can be filled in 8 ways (i,j,....p)
So number of ways they can be arranged is 16*12*8 = 4*4*4*3*4*2
This gives the number of arrangements where the order is important, but for us order is not important - It does not matter to us whether we select 'a' first of 'g' second... All that is important to us is that we have to select 3 people.
So we have to un-arrange the number of arrangements. We can do so by dividing the total number of arrangements with 3!.
So the final answer is 4*4*4*3*4*2/3*2 = 4^4.
Hence (B).
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