Question

How many ways are there to select 8 donuts from 6 types of donuts if at most 3 donuts are chosen from the first two types combined;
that is, at least 5 donuts chosen from the other four types. (Hint: break into cases)

answer: break into 4 cases for 0 or 1 or 2 or 3 in first two boxes
sum for i from 0 to 3 {C(2+i-1,i)xC(4+(8-i)-1,(8-i))};

Answer 1

The cases are given below and the number of ways:

Case	No. of donuts from first two types		No. of donuts from rest 4 types		Total ways
1	0	1	8	C(8+4-1,4-1) = C(8+3,3)	C(0,1)*C(11,3)
2	1	C(1+2-1,2-1) = C(1,1)	7	C(7+3,3)	C(1,1)*C(10,3)
3	2	C(2,1)	6	C(6+3,3)	C(2,1)*C(9,3)
4	3	C(3,1)	5	C(5+3,3)	C(3,1)*C(8,3)

To select p donuts from r types, it is equivalent to the number of non-negative solutions to the equation:

x1 + x2 + x3... xr = p

The number of ways: C(p + r - 1, r - 1)

The same formulas has been applied in the table above

It can be generalized into:

Number of ways = From 0 to 3: Σ C(i, 1) * C(11 - i, 3)