Confusion question in Permutation and combination    Q: How many ways can 4 prizes be given away to 3 students,
 if each boy is eligible for all the prizes?

Ans:
    Any one prize can be given to any one of the 3 students and
 hence there are 3 ways of distributing each prize.
    Hence, the 4 prizes can be distributed in 3^4= 81 ways.

My question is why can't  we solve in another way.
each student can be get one of 4 prizes. Then Answer will be 64.
Can somebody give a clear explanation on why second approach is wrong. 
__S1_____S2_________S3____
|__4___|__4_____|__4_____|

 A: " Each student can (only) be get given one of 4 prizes " is incorrect, because each student can be given 0, 1, 2, 3, or 4 prizes. Conclusions drawn from a false premise can be wrong, and that's the case here.
A: We have 4 prizes.
Prize 1: Okay, pick up the prize off of the prize table. Now look at each boy. You need to give the prize to one of them. Choose the boy you want to give the prize to and give it to him. You just chose between 3 options.
Prize 2: Okay, now pick up the 2nd prize off of the prize table. Now look at each boy. You need to give the prize to one of them. Choose the boy you want to give the prize to and give it to him. You just chose between 3 options. There were 3 previous options for where the 1st prize went. Thus we have $3 \cdot 3$ different ways two prizes could have been handed out
So on and so forth. We need to make 3 choices each time. This creates a trinary tree (A node is a decision, and edge is a prize given). The number of edges from depth $n$ to the root is the number of ways the prizes could be distributed.
