In how many ways can 20 identical chocolates be distributed among 8 students Provided each student gets atleast 1 chocolate and exactly two students get atleast two chocolates each.
We know that if k indistinguishable objects are to be placed in n bins such that each bin contains atleast 1 object- the # of ways we can do that is:
$\binom{k-1}{n-1}$
So here in this context, we have, $\binom{20-1}{8-1}=\binom{19}{7}$ as $k=20 \space identical \space chocolates$ and $n=8 \space distinct \space students$
Now after this step we have to find out which two students are given 2 chocolates each.
So choose two students: We have, $\binom{8}{2}$ # of ways to do that
Now after that we are left with $19-7=12$ chocolates which we have to divide among 2 students s.t each student gets atleast 2 chocolates.
Thus this boils down to $x+y=12,x\ge2,y\ge2$
So possible solutions:
$(2+10=12)....(occurs  \space twice,i.e 10+2=12),(3+9=12)....(occurs \space twice),(4+8=12)....(occurs \space twice),(5+7=12)....(occurs \space twice),(6+6=12)....(occurs \space only \space once)$
Thus total # of cases= $9$
So, for each of  $\binom{8}{2}$ students we have 9 cases.
So total # of instances=  $\binom{8}{2}\times 9$
Hence total # of instances=$\binom{19}{7}\times\binom{8}{2}\times 9$ 
But sadly this does not match with any of the options given:
Options are
A.$308$
B.$364$
C.$616$
D.$\binom{8}{2}\binom{17}{7}$
Where am I wrong?
What is/are the correct step(s)?
Please give proper and detailed reasoning.
My assumptions are wrong. Please see my answer which was suggested by @antkam to solve this.
 A: See @antkam helped me solve this. It is very very simple. Need not have taken all that hassle.
Ok Now see that 2 students are supposed to get atleast 2 chocolates each but there is no restriction with the other 6 students. They are all getting atleast 1 chocolate each.
Hence let us first allow all those 6 students get 1 chocolate each and this is only 1 case as the chocolates are identical.
Now # of chocolates left = $20-6=14$
We now have to distribute these 14 chocolates between the 2 students s.t each gets atleast 2 chocolates.
So this boils down to the following equation:
$x+y=14,x\ge2,y\ge2$
The possible solutions are:
$(2+12=14)....(counted \space twice),(3+11=14)....(counted \space twice),(4+10=14)....(counted \space twice),(5+9=14)....(counted \space twice),(6+8=14)....(counted \space twice),(7+7=14)....(counted \space **once**)$
Thus total $\#$ of cases = $2\times 5+1=11$
And in how many ways can we choose these $x \space \text{and}\space  y$?
$\#$ of ways we can choose $x \space \text{and}\space  y$= $\binom{8}{2}$
Thus total no. of cases= $11\times\binom{8}{2}\times1(\text{for the case of giving 6 students 1 chocolate each})=308$...(option A)
A: All $8$ students shall obtain $\geq1$ pieces, and exactly $2$ of them shall obtain $\geq2$ pieces. 
Give each student $1$ piece, select the $2$ special students in ${8\choose2}=28$ ways, give the senior of these $s\in[11]$ additional pieces, and the junior the remaining $12-s>0$ pieces. It follows that there are $28\cdot11=308$ admissible allocations in all.
