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.

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*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. 


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*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.
