Birthday problem involving overlapping birthday dates This is the last problem on my Combinatorics assignment.  The professor hasn't given us any example similar to this so I was wondering if you guys could help me!
How many people in a group do you need so that there is at least a 50% chance two people have

(a) a birthday in the same month?  (Assume all months have equal length.)

(b) a birthday in the same week? (Assume exactly 52 weeks in a year.) 

 A: $P$(atleast $2$ people having birthday in same month)
$=  
1−P$(no two people have birthday in same month)
$=1−\frac{12.11...(12-n+1)}{12^n}  
\geq1/2$ i.e. $ \frac{12.11...(12-n+1)}{12^n}\leq 1/2$  
So, you know as $n$ increases this probability will increase, and $n \leq 12$.
 For $n=4$, $P=\frac{12.11.10.9}{12.12.12.12}=\frac{990}{12^3}=\frac{990}{1728}\approx0.6$(you don't need exact)
Next will be $P_{n=5}=P_{n=4}*\frac8{12}<1/2$
Hence, $n=5$ for $(a)$, similarly try for (b)
A: These probems are very similar, I'll solve (a).
Let $p_n$ be the probability that $n$ guys have the same birthday in the same month.
You should know that $$p_n=1-\frac{\prod_{i=0}^{n-1}(12-i)}{12^n}=1-\prod_{i=0}^{n-1} \left(1-\frac{i}{12} \right)$$
we'll use the following inequality 

$$1-x \le e^{-x}$$

Valid $\forall x \in \mathbb{R}$ (why?). Once you apply it n times to $p_n$ expression you get
$$\prod_{i=0}^{n-1} \left(1-\frac{i}{12} \right) \le \prod_{i=0}^{n-1} e^{-i/12}=\exp{\left(-\sum_{i=0}^{n-1} \frac{i}{12} \right)}$$ so
$$p_n \ge 1-\exp{\left(-\sum_{i=0}^{n-1} \frac{i}{12} \right)}=1-\exp{\left(-\frac{n(n-1)}{24} \right)}$$
Now the $n$ you are looking for is the lowest $n \in \mathbb{N}$ sucht that 
$$1-\exp{\left(-\frac{n(n-1)}{24} \right)} \ge \frac{1}{2}$$
solve this and get $n=5$.

This metod works for other question like "How many people in a group do you need so that there is at least a 50% chance two people have a birthday in the same day?" which would be difficult to check a guessed answer .
