Distributing amounts of different objects to distinct people I have the following exercise and I am wondering if the subsequent solution of mine is correct.

Alan, Peter, Steve, and Donald have a total of 11 chocolate bars that
  they are going to divide among themselves. They have 9 Snickers (S), 1
  Mars (M) and 1 Bounty (B). In how many different ways can they
  distribute the chocolates among themselves such that each person gets
  at least one chocolate?

Each person shall have at least one chocolate each, i.e. a total of 4 chocolates. Call these compulsory. There are 4 different ways to choose compulsory chocolates:


*

*$\{S, S, S, S\}$

*$\{S, S, S, M\}$

*$\{S, S, S, B\}$

*$\{S, S, M, B\}$


Case 1: There is only one way to distribute the four compulsory Snickers among the boys. Then there are $\binom{4}{1}$ ways of distributing the Mars and $\binom{4}{1}$ ways of distributing the Bounty. The remaing five Snickers can be distributed in $\binom{4+5-1}{5}$ ways using the equation $a+b+c+d = 5$. In total: $\binom{4}{1}$$\binom{4}{1}$$\binom{4+5-1}{5}$.
Case 2: There are $\binom{4}{1}$ ways to distribute the one compulsory Mars and one way to distribute the three compulsory Snickers. Then there are $\binom{4}{1}$ ways to distribute the non compulsory Bounty and $\binom{4+6-1}{6}$ ways to distribute the six non compulsory Snickers. In total: $\binom{4}{1}$$\binom{4}{1}$$\binom{4+6-1}{6}$.
Case 3: There are $\binom{4}{1}$ ways to distribute the one compulsory Bounty, $\binom{4}{1}$ ways to distribute the one non compulsory Mars and $\binom{4+6-1}{6}$ ways of distributing the six remaining Snickers. In total: $\binom{4}{1}$$\binom{4}{1}$$\binom{4+6-1}{1}$.
Case 4: There are $\binom{4}{1}$ ways of distributing the one compulsory Mars, then $\binom{3}{1}$ ways of distributing the compulsory Bounty, and one way of distributing the two compulsory Snickers. The remaining seven non compulsory Snickers can be distributed in $\binom{4+7-1}{7}$ different ways. In total: $\binom{4}{1}$$\binom{3}{1}$$\binom{4+7-1}{7}$.
All in all: $\binom{4}{1}$$\binom{4}{1}$$\binom{4+5-1}{5}$ + $\binom{4}{1}$$\binom{4}{1}$$\binom{4+6-1}{6}$ + $\binom{4}{1}$$\binom{4}{1}$$\binom{4+6-1}{1}$ + $\binom{4}{1}$$\binom{3}{1}$$\binom{4+7-1}{7}$.

You seem to get something like $3824$, which looks far too many: my answer is about half that.
So looking at possible issues with your answer: 


*

*In case 3 you seem to have changed a 6 into a 1, so case 3 does not give the same answer as case 2; I am unclear about whether this is deliberate or not 

*I think you may have double, triple or quadruple counted cases like $(SSSS, SSM, SB, SS)$  

*To avoid double counting, you might consider cases where the compulsory Mars or Bounty are not accompanied by any Snickers.  You would need to add $\{S,S,S,MB\}$ to your patterns and do the calculations where non-compulsory bars can only be given to people who already have a compulsory Snickers. 
