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))};
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)
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