Addendum just added to respond to questions from the Original Poster.
As other answers have indicated, while it is true that there are $\binom{11}{5}$ non-negative integer solutions to
$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 6$,
these solutions are not equally likely.
Let's consider $3$ such solutions: I will use the syntax $(a,b,c,d,e,f)$ to indicate that
$x_1 = a, x_2 = b, x_3 = c, x_4 = d, x_5 = e, x_6 = f.$
Solution 1: (3,2,1,0,0,0).
There are $\binom{10}{3}$ ways that $3$ of the 1st type of candy could be selected.
There are $\binom{10}{2}$ ways that $2$ of the 2nd type of candy could be selected.
There are $\binom{10}{1}$ ways that $1$ of the 3rd type of candy could be selected.
Computation equals $\binom{10}{3} \times \binom{10}{2} \times \binom{10}{1}.$
Solution 2: (2,2,2,0,0,0).
There are $\binom{10}{2}$ ways that $2$ of the 1st type of candy could be selected.
There are $\binom{10}{2}$ ways that $2$ of the 2nd type of candy could be selected.
There are $\binom{10}{2}$ ways that $2$ of the 3rd type of candy could be selected.
Computation equals $\binom{10}{2} \times \binom{10}{2} \times \binom{10}{2}.$
Solution 3: (1,1,1,1,1,1).
There are $\binom{10}{1}$ ways that $1$ of the 1st type of candy could be selected.
There are $\binom{10}{1}$ ways that $1$ of the 2nd type of candy could be selected.
There are $\binom{10}{1}$ ways that $1$ of the 3rd type of candy could be selected.
There are $\binom{10}{1}$ ways that $1$ of the 4th type of candy could be selected.
There are $\binom{10}{1}$ ways that $1$ of the 5th type of candy could be selected.
There are $\binom{10}{1}$ ways that $1$ of the 6th type of candy could be selected.
Computation equals $\binom{10}{1}^6.$
My solution:
Probability equals $$\frac{N\text{(umerator)}}{D\text{(enominator)}}$$
Where $D = \binom{60}{6} = $ the number of ways of selecting $6$ candies out of $60$. Here:
- The selection is without replacement.
- The order that the candies are selected is not deemed relevant.
Because of the second point above, the numerator, $N$ must be enumerated in a consistent manner. That means that when enumerating $N$, you must presume that the order that the candies where selected is irrelevant.
Suppose that the candies were numbered $1,2,\cdots,60$, where :
- Candies $1$ through $10$ were Milky Way.
- Candies $11$ through $20$ were Almond Joy.
- Candies $21$ through $30$ were Butterfinger.
- Candies $31$ through $40$ were Nestle Crunch.
- Candies $41$ through $50$ were Snickers.
- Candies $51$ through $60$ were Kit Kat.
There are $\binom{10}{1} = 10$ choices for which Milky Way candy is selected.
Independently, regardless of which Milky Way candy is selected, there are $\binom{10}{1} = 10$ choices for which Almond Joy candy is selected.
The same type of reasoning indicates that you will have $6$ independent factors to consider: that is, for each type of candy, there are $(10)$ choices for which candy is selected.
In other words, you have $10$ choices for which Milky Way is selected, $10$ choices for which Almond Joy is selected, $10$ choices for which Butterfinger is selected, and so forth.
Therefore, $N = \binom{10}{1}^6 = \left(10\right)^6.$
Final computation:
$$\frac{N}{D} = \frac{(10)^6}{\binom{60}{6}}.$$
Addendum
Responding to the OP's comment and the latest editing of his question.
I actually was looking for an explanation of why they are not equally...
I understand most answers are claiming that all integer solutions are not 'equally likely', this somehow presupposes that one choice of the probability measure is correct over the other, eg: I could have defined the probability measure such that each integer solution is equally likely. Is there any reason to chose one measure over the other?
In order to explore these issues, I will first try to examine in greater detail, each of the $\binom{11}{5} = 462$ solutions to the equation
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 6 \tag1 $$
$$~~: ~x_1, x_2, x_3, x_4, x_5, x_6 \in \Bbb{Z_{\geq 0}}. $$
I will partition the solutions into all of the possible formats. For each individual solution that corresponds to a specific format, I will determine how many distinct distributions are represented by this solution. Then the probability will be computed as
$$\frac{\text{# of satisfying distributions}}{\text{total # of possible distributions}}.$$
I will use the syntax $(a,b,c,d,e,f)$ to indicate that
$x_1 = a, x_2 = b, x_3 = c, x_4 = d, x_5 = e, x_6 = f.$
For a solution that follows format $k$, as detailed below, I will use the variable $t_k$ to denote how many of the $(462)$ solutions have format $k$.
I will then use the variable $w_k$ to denote how many different distributions are associated with each individual solution to (1) above that follows format $k$.
In effect, $w_k$, which denotes a number of distributions, will be used to represent a weight that is given to each of the solutions represented by format $k$. In effect, $w_k$ will represent the relative weight to be given to each of the $462$ solutions, based on how many distributions are associated with the specific solution.
This implies that the product $(t_k \times w_k)$ will represent all of the distributions possible that are represented by format $k$.
The following table examines each of the possible formats:
\begin{array}{| r | r | r | r |r |}
\hline
\text{#} & \text{Pattern} & t_k
& w_k & t_k \times w_k \\
\hline
1 & (6,0,0,0,0,0) & 6 & \binom{10}{6} & 1,260 \\
\hline
2 & (5,1,0,0,0,0) & 30 & \binom{10}{5} \times \binom{10}{1} & 75,600 \\
\hline
3 & (4,2,0,0,0,0) & 30 & \binom{10}{4} \times \binom{10}{2} & 283,500 \\
\hline
4 & (4,1,1,0,0,0) & 60 & \binom{10}{4} \times \binom{10}{1}^2 & 1,260,000 \\
\hline
5 & (3,3,0,0,0,0) & 15 & \binom{10}{3}^2 & 216,000 \\
\hline
6 & (3,2,1,0,0,0) & 120 & \binom{10}{3} \times \binom{10}{2} \times \binom{10}{1} & 6,480,000 \\
\hline
7 & (3,1,1,1,0,0) & 60 & \binom{10}{3} \times \binom{10}{1}^3 & 7,200,000 \\
\hline
8 & (2,2,2,0,0,0) & 20 & \binom{10}{2}^3 & 1,822,500 \\
\hline
9 & (2,2,1,1,0,0) & 90 & \binom{10}{2}^2 \times \binom{10}{1}^2 & 18,225,000 \\
\hline
10 & (2,1,1,1,1,0) & 30 & \binom{10}{2} \times \binom{10}{1}^4 & 13,500,000 \\
\hline
11 & (1,1,1,1,1,1) & 1 & \binom{10}{1}^6 & 1,000,000 \\
\hline
\text{sum} & & 462 & & 50,063,860 \\
\hline
\end{array}
Notice that $\sum_{k=1}^{11} t_k = 462 = \binom{11}{5}.$
Notice that $\sum_{k=1}^{11} (t_k \times w_k) = 50,063,860 = \binom{60}{6}.$
This means that in the above chart, all of the $\binom{11}{5}$ possible solutions are accounted for and all of the $\binom{60}{6}$ possible distributions are accounted for.
Then, the probability of a satisfying distribution is
$$\frac{t_{11} \times w_{11}}{\sum_{k=1}^{11} (t_k \times w_k)} = \frac{1,000,000}{50,063,860} = \frac{10^6}{\binom{60}{6}}.$$
For illustrative purposes, I will discuss Format # $7$ in greater detail.
There are $6$ choices for which of the $6$ variables will be set to the value $(3)$. Once this selection is made, there are $\binom{5}{3} = 10$ ways of selecting $3$ of the remaining $5$ variables, and then setting these variables to the value $(1)$.
Therefore, $t_7 = 6 \times 10 = 60.$
For each of the $60$ out of a possible $462$ solutions to the equation $x_1 + \cdots + x_6 = 6$, how many distributions are represented by each solution? To answer that, simply determine how many distributions are represented by the specific solution $(3,1,1,1,0,0)$.
This solution calls for $3$ candies to be selected from the $10$ candies represented by the candy type associated with the variable $x_1$. This can be done in $\binom{10}{3}$ ways.
Similarly, there are $\binom{10}{1}$ possible ways of selecting $1$ candy out of $10$, for each of the candy types represented by the variables $x_2, x_3, x_4.$
This explains the value of $w_7$, shown in the chart, and explains the product $(t_7 \times w_7)$ shown in the chart.
Best Answer
(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 $\binom{5}{4}$.