I need some intuition for an element of the following question:
The answer starts with this:
I would like to know how they get to 6 choose 2. If I write it out (1122, 1133, 1144, 1155, 1166, 2233, 2244,..., 5566) I get 15 possible digits too, but there must be an easier way to see this.
Also, is it logical that the question apparently leaves out double pairs like 1111 and 2222?
On
In total there are $6^5$ ways to display five digits, selected without bias but allowing repetition (ie: independently). Next we count the (equally-probable) favored outcomes.
First we choose the two digits for the pair, and the singleton, $\binom 6 2\binom 4 1$, then we choose which from the five dice to host them, $\binom 5 2\binom 3 2\binom 1 1$. These factors can be represented by a multinonials. Giving us the probability: $$\dfrac 1{6^5}\dbinom 6 {2,1,3} \dbinom 5{2,2,1}~=~\dfrac{5!^2}{6^4~3!~2!^3}$$
(NB: the "$3$" indicates the digits we do not choose.)
As you say, some might also consider four of a kind to be two pairs of the same number. In that case you should also consider full house (three and two of a kind) and yahtzee (five of a kind). So that would add three more terms: What are they?
$$\cdots + \dfrac 1{6^5}\binom 6 {\color{white}{1,1,4}} \binom 5 {\color{white}{4,1}}+\dfrac 1{6^5}\binom 6 {\color{white}{1,1,4}}\binom 5{\color{white}{3,2}}+\dfrac 1{6^5}\binom 6 1\binom 5 5$$
When you have $n$ objects to choose from, and you want to choose $r$, if the order doesn't matter, and repetition is not allowed, the answer is $\displaystyle{n\choose r}$.
If the order does matter, then you'd get $n(n-1)\cdots(n-r+1)=P(n,r)$ by the multiplication rule.
However, the order doesn't matter, so that means that every selection of $r$ objects appears many times -- $r!$ to be exact. Thus, you divide $P(n,r)$ by $r!$ to get the real number of possible choices. This is $\displaystyle{P(n,r)\over r!} = {n\choose r}$.