What are the no of ways in which we can place 7 apples in 5 containers, given neither apples nor containers are identical. What are the no of ways in which we can place $7$ apples in $5$ containers such that each container contains at least $1$ apple, given neither apples nor containers are identical
My attempt is as follows:
As containers are not identical, so let's enumerate them as $C_1,C_2,C_3,C_4,C_5$ and as apples are not identical, so let's enumerate them as $A_1,A_2,A_3,A_4,A_5,A_6,A_7$
Now let's try to fill one apple in all the $5$ containers.
No of ways to fill one apple in container $C_1=7$
No of ways to fill one apple in container $C_2=6$
No of ways to fill one apple in container $C_3=5$
No of ways to fill one apple in container $C_4=4$
No of ways to fill one apple in container $C_5=3$
Let's multiply all of them to get the no of ways in which we can fill $1$ apple in each of the container$=7\cdot6\cdot5\cdot4\cdot3=2520$
Now in each of the $2520$ ways, $2$ apples will be left at the end, now as all the containers are containing at least one apple, we can put the remaining $2$ apples in any of the containers. 
So no of ways to place remaining $2$ apples in any of the containers$=5\cdot5=25$
So $2520\cdot25=63000$ should be the answer.But actual answer is $11760$.
Where am I making the mistake. I tried to find it but didn't get any breakthroughs.
 A: *

*Case 1:


First choose which container will have 3 apples. That you can do on $5$ ways. Then choose 3 apples you will put in it, that you can do ${7\choose 3} $ ways. Now put in each of the remaining containers 1 apple, that you can do on $4!$ ways. So in total you have $5\cdot {7\choose 3}\cdot 4! = 4200$ ways in this case. 


*

*Case 2:


Chose which containers will have 2 apples. That you can don on ${5\choose 2} =10$ ways. Then choose 2 apples for first one and 2 apples for the second one, that you can do on ${7\choose 2}\cdot {5\choose 2}$ ways. Now put in each of the remaining containers 1 apple, that you can do on $3!$ ways. So in total you have $10\cdot {7\choose 2}\cdot {5\choose 2}\cdot 3!= 12600$ ways in this case. 
All together you have $   \boxed{16800}$ ways.
A: The Twelvefold Way is your friend.  The number of surjections from a set with $n$ members to a set with $r$ members is $$r!\{\,^n_r\}$$ where the bracked combination is the Stirling number of the second kind.  So you're looking for $5!\{\,^7_5\}=120\cdot140=16800$.

You went wrong by over-counting.  For instance, if you put choose to apple 1 in container A in the first phase and then choose to put apple 7 in container A in the final phase with the two leftover apples, then it is the same result as if you had chosen to put apple 7 in A in the first part and apple 1 in A in the final phase.  However, you counted it as two different arrangements.
A: It depends on how you "put" the apples in the bins, or well on which resulting distributions you are going to
consider "different".
After establishing that either the apples and bins are distinguishable (labeled), you are left - fundamentally - 
with another choice to do:    

whether or not to consider the order of the balls inside each bin.

Practically, one case corresponds to using transparent bins into which the apples stacks one over the other, then you consider
the bin with $[a_1,a_2]$ different from $[a_2,a_1]$.
The other case to flat large bins, where the position of each apple cannot be distinguished.
More rigorously:
 - in the first case (order considered) you are partitioning the set $\{1,2, \cdots, 7 \}$
into a list of $5$ sub-lists (without repetition), ex. $[3,2],[1,5,7],[],[6,4],[]]$;
 - in the second (order not considered), the partition is into a list of sub-sets, 
which may be or not also empty, ex. $[\{ 2,3 \} , \{ 1,5,7 \}, \{ \}, \{ 4,6 \}, \{ \} ]$.
Now, concerning the computations, we have.
1) order considered
We establish first the quantity of apples in each bin, which we can do in a number of ways equal to
$$
Q_{\,e}  = \left( \matrix{
  7 + 5 - 1 \cr 
  7 \cr}  \right) = 330\quad Q_{\,ne}  = \left( \matrix{
  7 - 1 \cr 
  5 - 1 \cr}  \right) = 15
$$
depending on whether we allow empty bins to be present ($Q_{\,e}$) or not ($Q_{\,n\, e}$).
These are obtained with the "stars&bars" approach, or better as the "weak" and "standard" compositions
of $7$ into exactly $5$ parts.
Then we permute the apples in $7!$ ways, and separate them according to the quantities defined above.
Therefore
$$
\eqalign{
  & N_{\,e}  = \left( \matrix{
  7 + 5 - 1 \cr 
  7 \cr}  \right)7! = 7^{\,\overline {\,5\,} }  = 1\,663\,200  \cr 
  & N_{\,n\,e}  = \left( \matrix{
  7 - 1 \cr 
  5 - 1 \cr}  \right)7! = {{6^{\,\underline {\,4\,} } \;7!} \over {4!}} = 7 \cdot 6 \cdot 5 \cdot 6^{\,\underline {\,4\,} }
  = 7^{\,\underline {\,5\,} }  \cdot \left( {5 \cdot 4 + 5 \cdot 2} \right) = 75\,600 \cr} 
$$

That means that you omitted the case in which, when the remaining two balls go into the same bin, you have the choice of which to 
  put above and below.

2) order not considered
The computations in this case are
$$
\eqalign{
  & N_{\,e}  = 5^{\,7}  = 78\,125  \cr 
  & N_{\,n\,e}  = 5!\left\{ \matrix{
  7 \cr 
  5 \cr}  \right\} = 16\,800 \cr} 
$$
as already provided in a precedent answer.
Concerning the number $11 \, 760$ I have no idea of what it relates to.
