Correct way find number of combinations with "at most" and "at least" restrictions? If I have 4 items to choose from A,B,C,D and need to choose a total of 10. However I must have at least 1 A and at least 2 B's, and A,B,C,D cannot be greater than 5. 
Am I solving the question to correct way?
My solution: 
$${A+B+C+D = 10
\\(A+1-5)+(B+2-5)+(C-5)+(D-5) = 10
\\A+B+C+D = 27}$$
therefore the solution is:
$30!/(27!3!)$ or ${}^{30}C_3$ ? 
 A: The way to make an unordered selection of 10 items from 4 categories with:


*

*With at least 1 A, 2B: count the integer solutions to $A+B+C+D=10-3$, which is ${}^{10}C_3$

*With at least 6 A, 2B: count the integer solutions to $A+B+C+D=10-8$, which is ${}^{5}C_3$.

*With at least 1 A, 6B: count the integer solutions to $A+B+C+D=10-7$, which is ${}^{6}C_3$

*With at least 1 A, 2B, 6C: count the integer solutions to $A+B+C+D=10-9$, which is ${}^{4}C_3$

*With at least 1 A, 2B, 6D: is again ${}^{4}C_3$

*Since we cannot select more than one category with more than five, that is all.


Now apply the principle of inclusion and exclusion.
A: Start with the $A$ and two $B$s you require.  Now you need to pick seven more with  at most four $A$s, three $B$s, five $C$, five $D$s.  Note that you cannot violate more than one of these conditions.  You can use stars and bars to find the unrestricted combinations and subtract the ones that don't work.  If there are six $C$s there is only one other, so there are three to subtract.
