At closing time at a Dunkin Donuts store they still have $10$ vanilla, $20$ custard, $24$ cinnamon, and $30$ chocolate donuts available. Donuts of the same kind are regarded as identical.
(a) How many ways are there to purchase $8$ donuts?
(b) How many ways are there to purchase $15$ donuts with at least three of each kind?
(c) How many ways are there to purchase $15$ donuts?
Part (a):
We have four different flavor donuts to choose from. We would like to purchase $8$ of them. So this is a stars and bars problem. We have $8$ stars and $3$ bars. So this is $\binom{8+3}{3}=\binom{11}{3}=165$ ways.
Part (b):
Step $1$: Put three of each kind in the box.
Now we need to purchase $3$ more donuts. This is again, stars and bars.
$\binom{3+3}{3}=\binom{6}{3}=20$ ways.
Part (c):
I know we need to be careful because there are only $10$ vanilla donuts left.
This will be a stars and bars problem.
How do I go about this problem?
The question can be formulated as: $$x_1+x_2+x_3+x_4=15,\\ 0\le x_1\le 10,\\ 0\le x_2\le 20,\\ 0\le x_3\le 24,\\ 0\le x_4\le 30,$$ which is equivalent to: $$x_1+x_2+x_3+x_4=15,\\ 0\le x_1\le 10,\\ 0\le x_2\le 15,\\ 0\le x_3\le 15,\\ 0\le x_4\le 15.$$ which is the difference between (1): $$x_1+x_2+x_3+x_4=15,\\ 0\le x_1\le 15,\\ 0\le x_2\le 15,\\ 0\le x_3\le 15,\\ 0\le x_4\le 15,$$ and (2): $$x_1+x_2+x_3+x_4=15,\\ 11\le x_1\le 15,\\ 0\le x_2\le 15,\\ 0\le x_3\le 15,\\ 0\le x_4\le 15,$$ (1):${15+4-1\choose 4-1}={18\choose 3}$.
(2): $$x_1+x_2+x_3+x_4=15,\\ 11\le x_1\le 15,\\ 0\le x_2\le 15,\\ 0\le x_3\le 15,\\ 0\le x_4\le 15,$$ which is equivalent to (let $x_1=y_1+11$): $$y_1+x_2+x_3+x_4=4,\\ 0\le y_1\le 4,\\ 0\le x_2\le 4,\\ 0\le x_3\le 4,\\ 0\le x_4\le 4,$$ which is: ${4+4-1\choose 4-1}={7\choose 3}$. Thus, the final answer is: ${18\choose 3}-{7\choose 3}=781$.