Why is the probability multiplied by $\binom{n}{k}$ A while ago I asked a question about probability here Why is binomial probability used here?
I get that you can find how many ways of choosing the $6$ correct out of $10$ questions.
But why do we multiply by $\binom{n}{k}$? 
I thought it is simple casework that:
$$P(\text{total probability}) = P(\text{Q 1->6 right and 7-10 wrong} + P(\text{Q 1->5 right, 6 wrong, 7 right, 8-10 wrong}) + ...$$
What is the idea? 
 A: You are correct that the simple casework will work. However, if you were to look carefully at the casework, you will notice something interesting. In each case, the actual probabilities are the same—you will have 6 correct and 4 incorrect answers. The only thing that changes is which of the 10 are correct or wrong.
So, you are looking for all of the patterns using just the letters "C" and "I" (for example) of length 10 where 6 are "C" and 4 are "I". Your cases becomes adding CCCCCCIIII and CCCCCIIIIC and CCCCIIIICC etc.
Guess what, the number of cases you will have to add individually are the number of ways to order 6 success in 10 events, or ${10 \choose 6}$. So all the multiplication is doing is saving you a whole bunch of additions. Well, that is really all what multiplication is—repeated application of addition. Since we know how many identical items we have to add, we multiply.
A: How many ways can you get 6 questions right?
1->6 right and 7->10 wrong is an event. But you need to count the others. For instance 1->3 wrong and 4->10 right. 
How many ways can you get 6 questions  out of 10 right? Choose $6$ out of $10$ to get right: ${10\choose 6}$. The rest will follow if you understand this.
A: $\binom{n}{k}$ is the number of ways of choosing $k$ things from $n$ things (without caring about the order), for example to get a $x^2$ term from (the binomial) $(1+x)^3=(1+x)(1+x)(1+x)$, I have to choose two $x$s but am free to choose which two, so I have $\binom{3}{2} = 3$ choices for this - I can discard any of the three 1s to choose two $x$s, giving $3x^2$ overall
