What is the probability that Jen grabs exactly five varieties of chocolate candy? 
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.
(a)  What is the probability that she got one of each variety?
(b)  What is the probability that Jen grabs exactly five varieties?

Part (a) was pretty straight forward and I had no problems.  Part (b) is where I was struggling.  I understood that the terms $\binom{10}{1}^4$ and $\binom{10}{2}$ has to be in the numerator of calculating the probability.  This is because out of the 5 types of candy selected, 4/6 types of candy had 1/10 candies selected.  I just wasn't sure how to represent  the different combinations of the 5 types of candies selected.
In other words, it was clear that the total combinations is $\binom{60}{6}$, but how do you express the numerator or number of favorables?
 A: (a)  $\cfrac{\binom{10}{1}^6\binom{6}{6}}{\binom{60}{6}}=0.01997$.  For each type of candy, there's 10 items, and you choose 1.  You do this six times.
(b)  In this scenario, we choose 4 of 6 types of candy and in each of those we will choose 1 from 10.  From the remaining 2 of 6 types of candy, we will choose one type of candy and that one will choose 2 from 10.  So the solution is:
$\cfrac{\left[\binom{6}{4}\binom{2}{1}\right]\left[\binom{10}{1}^4\binom{10}{2}\right]}{\binom{60}{6}}=0.2697$.
Where:
$\binom{6}{4}\binom{2}{1}=\binom{6}{4,1}$ is the # of ways to choose 5 candies from 6 (assign the pair last).
$\binom{10}{1}^4\binom{10}{2}:$ within each of those 5 candies, there are respective possibilities to choose them (Multiplication principle)

Alternatively, we could view this problem in a different but equivalent manner:  
$\cfrac{\left[\binom{6}{1}\binom{5}{4}\right]\left[\binom{10}{1}^4\binom{10}{2}\right]}{\binom{60}{6}}=0.2697$.  
$\binom{6}{1}\binom{5}{4}=\binom{6}{4,1}$  First choose the pair then choose the 4 others.
There are six slots we must occupy, but two of those slots will be the same candy.  If we begin to fill the slot which has 2 of the same candy, then we have $\binom{6}{1}$ ways of filling that slot.  Once that slot is occupied, there are 5 types of candy remaining with 4 slots to fill.  This gives us $\binom{5}{4}$.
