Counting Donut Problem 
We are going to select $12$ donuts from among sprinkles, caramel, chocolate, and maple (all equally likely). What is the probability that we select exactly $3$ maple and $1 $ chocolate donuts? 

This means that there are three partitions, so total number of spots would be $15.$ I then figure that ${15\choose 12}$ would give all possible combinations but how do I get to the point of $3$ maple and 1 chocolate?
Final ans given: $9/455$
 A: We assume that we select the doughnuts one at a time, and that the  $4^{12}$ possible sequences of choices are all  equally likely.  
The number of favourable sequences is $\binom{12}{3}\binom{9}{1}2^{8}$. For we are counting the $12$-letter words  in a $4$-letter alphabet A, B, C, D that have $3$ A's and $1$ B. The locations of the A's can be chosen in $\binom{12}{3}$ ways, and for each way the location of the B can be chosen in $\binom{9}{1}$ ways. The remaining $8$ spots can be filled with C's and/or D's in $2^8$ ways. 
For the probability, divide the number of favourables by the total number of possibilities.  
Remark: I prefer the following approach. Any particular sequence that has $3$ maple (it really should be "maple," as in oil-derived artificial maple, perhaps heavy oil, to give it a Canadian flavour), $1$ chocolate, and $8$ other has probability $(1/4)^3(1/4)^1(2/4)^8$, and there are $\binom{12}{3,1,8}$ (multinomial coefficient) such sequences. So the probability is $\binom{12}{3,1,8}(1/4)^3(1/4)^1(2/4)^8$.
A: Note: you are selecting a grand total of 12 donuts.
The total permutations (where order matters) for your choices number 4^12. 
Also consider: Since picking any given flavor is exactly as likely as picking any other, all possible overall permutations occur with the same probability.
However, what you are looking for is a bit more specific. You want, in your selection, to have 3 maples and 1 chocolate, precisely! The rest of the choices can be any combination of caramel and sprinkles. So how many ways can this occur?
Well you have 12 objects, 3 identical (maples), 1 specified (chocolate), and 8 that can be as they please from 2 options. 
So we have the multinomial $12!/3!1!8! that represents the number of arrangements for 3 maples, 1 chocolate and 8 "abstract scoops". However, these "abstract scoops", when counting permutations, account for many different scoop selections, in fact there will be 2^8 different selections (where order counts). For the maples and chocolate, well there is only one way you can pick 3 maples and 1 chocolate, precisely by picking, so 1 (or 1^1*1^3 if you want to be very rigorous :) ).
So the total number of permutations for what you want is: $(12!/(3!*1!*8!))*2^8$
So the probability that you will get such a permutation is the quotient of that and 4^12
If you aren't sure about the above, check each individual case, where your constants are 1 chocolate scoop and 3 maples scoops. So a given scenario to check would be: 
1 choc, 3 maple, 1 vanilla, 7 caramel or $12!/1!3!1!7!$
You will find that it all sums to (12!/(3!*1!*8!))*2^8$
This beautiful property is due to the nice identity:
$$2^n = n!/0!n! + n!/1!(n-1)! + n!/2!(n-2)! + n!/3!(n-3)! + … + n!/n!(n-n)!$$
I.e. a row in pascal's triangle is equal to 2^(number of the row). 
A: We make the assumption that all unordered selections of 12 doughnuts are equally likely (so that, for example, choosing 12 maple is just as likely as choosing 4 maple, 4 chocolate, and 4 caramel).
If we choose exactly 3 maple and 1 chocolate, then there are only 9 ways to select the other 8 doughnuts since the number n of possibilities for the caramel satisfies $0\le n\le8$ 
(or using $\binom{9}{1}$, since there are 8 doughnuts and one divider needed).
The total number of ways to choose the 12 doughnuts is given by $\binom{15}{12}=\binom{15}{3}=455$, since this is equal to the number of ways to arrange 12 dots and 3 dividers.
Therefore the probability (under the assumption stated above) is given by $\frac{9}{455}$.
A: You are counting the number of ways to arrange the letters MMMCOOOOOOOO, $12!/\left(3!1!8!\right)$, where M=Maple, C=Chocolate and O=Other and each selection has a probability of P(M) = P(C) = 1/4 and P(O) = 1/2 so for any particular arrangement the probability is $\left(\frac{1}{4}\right)^{4}\left(\frac{1}{2}\right)^{8}$. The total probability is just the product.
