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.
Best Answer
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.