Probability of at least three students having the same zodiac sign Q: 

There are $30$ students in a class. What is the probability that at least three of these
  students have the same zodiac sign ($12$ signs)

I put the two variables in the question into sets $A$, $B$ and noted that their cardinalities:
$|A| = 30$, $|B| = 12$
And by the rule of counting functions I found that the number of ways of assigning a student a star sign to be $12^{30}$.
However I'm unsure what to use as the numerator and if this is the correct total to use for the denominator?
 A: You could assign at most a group of 24 students to have exactly 2 signs in common. That means the 25th student must be given a sign already assigned to two other students. 
Thus the probability of at least 3 students from a group of 30 having the same sign is 1, since it is unavoidable.
This called the pigeonhole principle and is a perfectly legitimate "mathematical" way to prove something.
UPDATE: Suppose there were only 20 students in the class. Then there are ways to assign zodiac signs to the students so that no sign is shared by more than 2 students. How many is the big question. 
Think of filling 12 distinguishable bins (the zodiac signs) with 20 distinguishable balls (the students):
1.) I could fill 10 bins with 2 balls each, leaving 2 bins empty. There are $\binom{12}{10}$ ways to select the bins I fill and each choice can be filled $\frac{20!}{2!^{10}}$. Thus there are  $$\binom{12}{10}\cdot \frac{20!}{2^{10}}$$ ways I could fill 10 bins with 2 balls each.
2.) I could fill 9 bins with 2 balls each and 2 bins with 1 ball each, leaving 1 bin empty. There are $\binom{12}{9}$ ways to select the bins I fill with 2 balls and $\binom{3}{2}$ ways to choose the two bin I put 1 ball in. Each choice of 2 bins can be filled $\frac{20!}{2!}\frac{1}{2!^{9}}=\frac{20!}{2!^{10}}$ ways and each choice of single bins can be filled $2!$ ways. Thus there are  $$\binom{12}{9}\cdot \frac{20!}{2!^{10}}\cdot\binom{3}{2}\cdot 2!=\binom{12}{9}\cdot\binom{3}{2}\cdot\frac{20!}{2^{9}}$$ ways I could fill 9 bins with 2 balls each and 2 bins with 1 ball each, leaving 1 bin empty.
3.) I could fill 8 bins with 2 balls each and 4 bins with 1 ball each, leaving no bins empty. There are $\binom{12}{8}$ ways to select the bins I fill with 2 balls and with all the rest getting single balls. Each choice of 2 bins can be filled $\frac{20!}{4!}\frac{1}{2!^{8}}=\frac{20!}{4!\cdot 2!^{8}}$ ways and each choice of single bins can be filled $4!$ ways. Thus there are  $$\binom{12}{8}\cdot\frac{20!}{4!\cdot 2!^{8}}\cdot 4!=\binom{12}{8}\cdot\frac{20!}{2^{8}}$$ ways I could fill 9 bins with 2 balls each and 2 bins with 1 ball each, leaving 1 bin empty.
Thus the probability of any random arrangement have 3 or more students with the same zodiac sign is
$$
\frac{12^{20}-\left(\binom{12}{10}\cdot\frac{20!}{2^{10}}+\binom{12}{9}\cdot\binom{3}{2}\cdot\frac{20!}{2^{9}}+\binom{12}{8}\cdot\frac{20!}{2^{8}}\right)}{12^{20}}.
$$
A: Your choice of denominator is correct.
Firstly, if one looks at the problem, one may realize that this must occur.  Given twelve possible choices of zodiac sign, the maximum class size possible where only two or fewer students have the same zodiac sign is 12*2=24.  With a class larger than 24, every possible permutation must satisfy the given condition, resulting in a probability greater than or equal to 1.  There is no possible way to not satisfy the condition.
To show this mathematically:
**THE FOLLOWING METHOD IS INCORRECT, VIEW LAARS' ANSWER FOR MORE INFORMATION **
Given your choice of denominator, all you have do for the numerator is count the total number of ways that any three students could have the same sign.  Your logic should follow as such:


*

*Choose which sign the three students have

*Choose which three students have the chosen sign

*It does not matter what the sign of the remaining 27 students have.
Number of possible signs: 12
Number of possible ways to choose three students: 30 choose 3
Number of possible ways to choose signs for the remaining 27 students: 12^27
Therefore your numerator should be
12 * (30 choose 3) * (12^27)
The value of this is greater than your denominator, resulting in a probability greater than one.  This is due to the fact that no permutation exists in which the given condition does not occur.  The condition is guaranteed to be satisfied always.
