Different Perspectives of Multinomial Theorem & Partitions There are 2 important interpretations of the multinomial theorem and coefficients.
1:  Determining the number of ordered strings that can be formed using a set of letters.  For example, with 1 m, 4 i's, 4 s's , and 2 p's, there are $\binom{11}{1,4,4,2}=34,650$ possible 11-letter strings that can be formed, of which mississippi is one specific example.
2: Partitioning a set of objects into several groupings where objects in each grouping are indistinguishable.  For example, in a group of 11 candidates, how many ways can we form 4 committees such that one committee has only 1 member, another has 4 members, another has 4 members, and the fourth committee has 2 members (assuming each person can serve in one committee)?  Again the answer is $\binom{11}{1,4,4,2}=34,650$.  I also understand that multinomial coefficients are a shortcut to partitioning.  We could have equivalently written:  $\binom{11}{1}\binom{10}{4}\binom{6}{4}\binom{2}{2}=34,650$.
I did one problem the other day (Probability/Combinatorics Question), and answered it.  However, I am reviewing the material and seem to get the concepts confused.  In the following question:

At a picnic, there was a bowl of chocolate candy that had 10 pieces of Milky Way, Almond Joy, Butterfinger, Nestle Crunch, Snickers, and Kit Kat. Jen grabbed six pieces at random from this bowl of 60 chocolate candies.
(b) What is the probability that Jen grabs exactly five varieties?

When calculating the numerator of the probability, sometimes I am tempted to write $\binom{6}{2,1,1,1,1,0}$ instead of the correct $\binom{6}{4,1}$.  Sometimes I want to view a type of candy as a committee group.  So since we have 5 varieties, we have 5 committees where one committee has 2 people (or candies), and the rest of the 4 committees have 1 person (or 1 candy).  Can anyone please explain why this is the wrong view and how I can stop confusing myself? 
Thank you in advance.
 A: If you want to know why your reasoning is flawed, by specifying the $6$ candies as committees and assigning the 6 candies (or people) that Jane grabs to committees in the following way:
$2$ people for $c_1$, $1$ person for each of $c_2,c_3,c_4,c_5$, and $0$ people for $c_6$ 
you are fixing the committees that accept $2$ people and $0$ people. In reality any committee is eligible to be the committee that takes $2$ people (or the committee that takes $0$ people). Your model is counting the number ways to distribute the $6$ people given that some fixed committee (say, always Milky Way) takes $2$ people, and another committee (say, always Kit Kat), takes $0$ people, and the rest take $1$ person.
A: A better way to solve this problem and to evade the confusion that arises is to solve it in 2 parts, so to speak. The first part, will be finding the denominator and this is straight forward, the number of ways in which one can choose 6 out of 60.
The second part, where the numerator is to be determined, we should resort to finding out the number of ways In which we can pick up 5 varieties and one out of the 55 that remain.
This is not as straightforward as the denominator but we can consider the fact that we have complete control over the process and it won't take much time before you realise the numerator is
10^5*55
And the probability comes out to be ~0.1
