Distributing 3 objects in 5 boxes In how many ways can 3 objects be distributed in 5 boxes so that no 2 objects go to the same box.
My reasoning is as follows:
Since, no 2 objects can be in the same box, each box possess a unique object, so only 3 boxes can be acquired at once.  Arranging 5 boxes with 3 at once is 5.4.3 = 60 ways.
But notice, for each of that 60 way the 3 objects can be arranged in 3! Ways.
Thus, total ways = 60.3!
But the book argues as follows:
Since there is a 1-1 correspondence between distributions of 3 objects into 5 boxes with the condition stated. Thus, the required number of such distributions is p(5,3)= 60 ways
What are the gaps in my reasoning. What am I missing?
 A: Imagine that there are just $3$ boxes and $3$ balls, each labelled, the boxes A, B, C and the balls 1,2,3.
How many ways can we arrange these objects with one ball into each box?
It's just $6$. You have claimed the answer is $36$, because we can re-arrange the boxes $3!=6$ ways and also we can re-arrange the balls in $3!=6$ for a total $36$ different arrangements.
But they are not all different. If we have boxes BCA and balls 123, this is the same as ABC 312 or CBA 213.
So we fix one of the set of objects, say the boxes, as ABC, and place the balls into them, and we can do this $6$ ways ($123,132,213,231,312,321$).
Then we choose $3$ boxes from $5$, and there are $\binom53=10$ ways to do this. Fix the particular set of boxes, and there are $6$ ways to place the balls, and using the principle of multiplication, there are $10\cdot6=60$ ways in total.
A: For the first object there are 5 different boxes and for the second one there are 4 different boxes and for the third one 3 different boxes.(no two of them in the same box)
By the multiplication principle $5\times4\times3=60$
Or
$P(5;3)=60$
