dividing objects in combinatorics so I have this question with a lot of sub questions inside, i'm sure about my answer for most of them, still i have 2 that i dont really know how to do :
There are 6 girls that robbed a bank, stole 100 coins and they need to divide them between them, in how many ways they can split them so :
a) 1 spesific girl, lets call her Jessica, can't get more than 10 coins.
b) none of the 6 girls will get more than 50 coins.
for a, i know that what you do in these situation is you say that the answer is equal to the answer of x1 + x2 + x3 + x4 + x5 + x6 = 100, but x1 has to be lower than 10 so you change it to y - 10  x2 + x3 + x4 + x5 +x6 = 100, move the 10 to the other side and you get 90... than you split it between 6. i'm not sure about it because the minus before the y bugs me. as for b, If this is the right way to do it i'll get 205 chooses 200 (if i change the x to 50 - y), still not sure.
Thank you in advance.
 A: This is effectively selection from a multiset. We are selecting 100 from a multiset with limits of {10,50,50,50,50,50}.
The process to follow is simplified in this case because the limits for most of the possibilities are the same. It follows an inclusion-exclusion type rule.
First establish the quantity for no limits. This is a "stars&bars" category selection:  the answer here is $M={105\choose 5}$
Then for each of the participants find the quantity of these selections that break their constraint. For Jessica she needs a minimum of 11 to break her constraint; allocate 11 to Jessica and repeat the free allocation, $B_1 = {94\choose 5}$ For the other robbers with the constraint-breaking 51 pre-allocated it is $B_i={54\choose 5}, i=2..6$
Now establish pairs of participants simultaneously breaking constraints. Fortunately only Jessica and any one other can do this. $C_{1i} = {43\choose 5}, i=2..6$
There are no triples that can simultaneously break constraint.
Using the inclusion-exclusion principle, the answer $T$ is then $T=M-\sum_i Bi + \sum_{i,j} C_{ij}$
$$T={105\choose 5} - {94\choose 5} -5{54\choose 5} +5{43\choose 5}$$


The above arises from considering both conditions (a) and (b) to hold. Re-reading, it looks like those might have been separate questions, but the process is good although the answers are simpler.
for only condition (a), Jessica (the driver?) only gets at most 10, no other constraints, the answer is $T_a = M-B_1$ (as defined above) 
$$T_a = {105\choose 5} - {94\choose 5}$$
for only condition (b), max payout to anyone is 50, the answer is $T_b= M-6B_2$ (as defined above), since there is no potential for breaking the payout limit on two of the gang simultaneously.
$$T_b = {105\choose 5} - 6{54\choose 5}$$
