How many ways to select a subset of eight doughnuts from three types of doughnuts if at most two doughnuts of the first type can be chosen? Please check my answer
I start by find all possible case that we can choose doughnuts without concern the condition 
$$\binom{(8+3-1)}{8} = \binom{10}{8}$$
then  I consider the case that fit the condition
Um,I'm not sure about find directly I choose to find by only the ways to doughtnuts by exclude fist type instead  that is
$$\binom{(8+2-1)}{8} = \binom{9}{8}$$
then the set of th way to pick doughtnuts that fit the condition is 
$$\binom{10}{8}-\binom{9}{8}$$
 A: The first part is ok, but to continue, you should write it as $\binom{10}2$ rather than $\binom{10}8$
Now to exclude cases with more than $2$ of the first type, we make sure we violate the given condition by choosing $3$ of the first type right at the start,
and choose only $5$ more doughnuts in any which way, thus $\binom{5+3-1}{3-1} = \binom72$
The final answer is thus $\binom{10}2 - \binom72$
A: When we take 0 doughnuts of first type , we can simply say that the sum of number of doughnuts of other two types should be 8. So there are 9 ways possible here in case 1. 
ex: If A and B are the other two types, we can assign values as number of doughnuts like if A=1, B must be 7 because the sum must be eight. We get 9 possible combinations for (A,B)
Next if we take 1 doughnut of first type, the sum of of other two types should be 7. So, 7+1 =8 possible combinations here. Next if we take 2 of first type, we get 7 possible combinations. So the total number of combinations is 7+8+9 = 24 . Tell me if I am wrong.
