Probability of occurring 
The numerial key for a specific safe is four digits long. The digits must be
  numbers from $1$ to $6$. (an example will be 4321 or 2564)
a) I threw a die randomly $4$ times. What will be the probability that
  I open the safe? How about repeating the same process $3$ times?
At a school in Utah there is a bench that can fit 2 people. We took an
  exam  and concluded that it is impossible to copy from someone that is
  not sitting in the same bench as you. So, the teacher gave assigned
  sits which were at random. The class consists of twenty students and ten benches.
  We have that there are 12 A's and 8 B's. Now we have that 8 students
  are at the same bench where both have A’s and 6 students that has B's
  sitting at the same bench. What is the probability that this will occur?

My attempt at the solution:
A die contains 6 digits and each digit has a $\frac{1}{6}$ chance of occuring so there is a possible of $6^4$ ways of rolling the dice four times. To get a possible of just one way of arrangments of tosses is just 1. So the probaility of getting the correct code to open the safe then will be $\frac{1}{6^4}$. If we repeat the process three times then it will be $3\dot\ \frac{1}{6^4}$. Is that right?
For the second question I do not know what to do because I don't even understand the question.
 A: For the first problem, think of the answer this way. The probability of having at least one set of dice rolls open up the safe is the same as the probability of not having no set of dice that has the safe combination (note the double negative in this sentence). The probability that a single set of dice doesn't open up the safe is $$\frac{6^4-1}{6^4} = \frac{1295}{1296}$$
So the probabilty that 3 dice rolls don't open up the safe is $$(\frac{1295}{1296})^3=0.99769$$
The probability that at least one of the sets of die rolls of the 3 open up the safe is the compliment of the answer we just got, so $1-0.99769 = 0.002313$ which is your final answer.
For clarification on your second question, just imagine 12 A's and 8 B's all in a line. Then, they were randomly put in pairs. What is the probability that from the 10 pairs that were formed, 4 pairs had both A's, 3 pairs had both B's, and since 4 A's are left and 2 B's are left, an additional pair of A's must be formed, and the last two pairs can either be AA,BB or AB,AB.
I am actually not too sure how to solve the second one, but this is what I would try. The probability of the said event occurring is
$$P(X)=\frac{\text{Number of ways 10 pairs can be made such that 4 have 2 A's and 3 have 2 B's}}{\text{Number of different pairs that can possibily be made}}$$
Which we will say equals $\frac ET$.
$T$ is the easiest to calculuate. Imagine that the way the instructor made the pairs was to rearrange all 20 people in a line randomly and then split them off by pairs down the line. That means that $20!$ pairs can be made. But wait, in those $20!$ pairs we have so many repeats! For instance, each person in each pair can be interchangeable, so we divide $20!$ by 2 for every pair we have, or we divide it by $2^{10}$. Also, it does not matter how we arrange the pairs themselves, so we must divide what we have so far by the factorial of the number of pairs, or $10!$. That means that $$T=\frac{20!}{(2^{10})(10!)}$$
We'll leave it like that for right now. Now we can work on $E$. We know that 4 pairs have both A's and 3 pairs have both B's. Suppose the instructor arranged the line such that it fit the following grade pattern: 
AA|AA|AA|AA|AA|BB|BB|BB|XX|XX

Hopefully you can go on from here...I am a bit busy right now and will edit this answer later if you still do not understand.
