a) Here a dozen items must come from a single variety. And we have 20 varieties possible. Hence the no of ways of selecting a dozen of same variety donut = 20.
b) For this part lets find out no of ways in which the 12 donuts can be chosen from 20 varieties
$$x_1 + x_2 + \dots +x_{20}=12$$
No ways of solving this is ${}^{(20+12-1)}C_{19} =141,120,525$ (Explained at end). In the question they have asked that there are at least 2 varieties of donuts. Hence we need to negate the ans of part a(i.e, the no of ways in which the same variety can be chosen ) from the total no of ways of choosing the donuts $= 141,120,525-20=141,120,505$
c) It is given that at least 6 donuts of blueberry variety should be there. Here we can assume that 6 donuts are already chosen. Hence now the solution is ${}^{(20+6-1)}C_{19}($ ie., no ways of solving $x_1+x_2+ \dots+x_{20}=6) =177,100$.
d) Here the condition is no more than 6 blue berry are selected. To solve this find the no of ways in which at least 7 blue berry will be chosen an in part c which is ${}^{(20+5-1)}C_{19}=42505$. Now negate this from total no of possibilities to get the required ans $= 141,120,525 - 42,505 = 141,078,021$.
We have $x_1 + x_2 + \dots + x_{20} = 12$.
So, this problem is equivalent to dividing 12 identical balls into 20 distinct bins without any further restrictions. So, lets use | for the separation between bins and $0$ for the balls. For example the below configuration shows all 12 balls in the first bin.
0 0 0 0 0 0 0 0 0 0 0 0 | | | | | | | | | | | | | | | | | | |
So, now our required answer will be all possible permutations of the above. We have 12 balls and (20-1) separations. Further 12 balls are identical and 19 separations are also identical (bins are distinct but separations are identical as two separations together means a bin in empty and their order doesn't matter). So, no. of possible permutations
$ = \frac{ (12+ 19)!}{12!19!} = {}^{31}C_{19}$.
This can be extended for $n$ bins and $k$ balls as ${}^{n+k-1}C_{n-1}$
Ref:http://www.cse.iitm.ac.in/~theory/tcslab/mfcs98page/mfcshtml/notes1/thperset.html
