Probability that in a group of 25 people there are at least 10 unique birth months I am trying to find the probability that in a group of 25 people there are at least 10 unique birth months using the principle of inclusion exclusion.
Principle of inclusion exclusion
$\bigcup_{i=1}^{n} A_{i}\left|=\sum_{i=1}^{n}\right| A_{i}\left|-\sum_{1 \leq i<j \leq n}\right| A_{i} \cap A_{j}\left|+\sum_{1 \leq i<j<k \leq n}\right| A_{i} \cap A_{j} \cap A_{k}\left|-\cdots+(-1)^{n-1}\right| A_{1} \cap \cdots \cap A_{n} \mid$
I approached the problem as follows.
Let $M_1,... M_{12}$ represent the months January, ..., December.
Let $M_k^{c}$ be the event that no one is born in $M_k$. Then $P(M_k)^{c} = (11/12)^{25}$. Similarly, the probability that no one is born in any of x months is $((12-x)/12)^{25}$.
Using the principle of inclusion exclusion,
$$P(M_{1}^{c} U M_{2}^{c} U ... U M_{10}^{c}) = {10 \choose 1}(11/12)^{25} - {10 \choose 2}(10/12)^{25} +{10 \choose 3}(9/12)^{25} - {10 \choose 4}(8/12)^{25} + ... -{10 \choose 10}(2/12)^{25}$$
The probability that at least one person is born in each of the months $M_{1}$ through $M_{10}$ is 1 - $P(M_{1}^{c} U M_{2}^{c} U ... U M_{10}^{c})$.
I wrote the following Matlab code to evaluate this expression. The output is approximately 0.25.
fprintf("%.60f\n",findp2(25)) 
function output = findp2(p)
    %the output is the probability that there are at least 10 unique 
    %birthmonth among p people 
pc = 0; 
for k = 1:10 
pc = pc + (-1)^(k-1)*nchoosek(10,k)*((12-k)/12)^(p);  
end 
format long  
output = 1-pc;   
end

Because this is the probability for 10 specific months, I multiplied this by ${12 \choose 2}$ to obtain 16.5 as the probability that there are at least 10 unique birth months among 25 people.
Can someone explain what I am doing wrong?
 A: "Can someone explain what I am doing wrong?"  You correctly applied inclusion-exclusion for the first part when talking about the probability at least one of the months Jan-Oct were missing.  You then completely ignored inclusion-exclusion when talking about the probability there was at least some group of ten months who all appeared, having just multiplied by $\binom{12}{2}$.  You should have then applied inclusion-exclusion on this as well, subtracting $12$ times the probability that some eleven months all appeared and then adding the probability that all twelve months appeared.

As for a more streamlined calculation (but requires stronger machinery), consider using Stirling Numbers of the Second Kind.  There are ${{25}\brace{10}}\cdot 12\frac{10}{~}$ ways to break $25$ people into $10$ nonempty groups and assign a month to each of those groups.  Similarly for $11$ months it would have been ${{25}\brace{11}}\cdot 12\frac{11}{~}$ and all $12$ months it would have been ${{25}\brace{12}}\cdot 12!$.  Adding these and then dividing the result by $12^{25}$ completes the problem, giving:
$$\dfrac{{{25}\brace{10}}\cdot 12\frac{10}{~}+{{25}\brace{11}}\cdot 12\frac{11}{~}+{{25}\brace{12}}\cdot 12!}{12^{25}}\approx 0.886841\dots$$
Notation notes:  ${n\brace k}$ represents the Stirling Number of the Second Kind denoting the number of ways $n$ distinct objects can be partitioned into $k$ unlabeled nonempty parts.  $n\frac{k}{~}$ is another way of writing the falling factorial $\frac{n!}{(n-k)!}$ and denotes the number of injective functions from a set with $k$ elements to a set with $n$ elements.
A: Using PIE
Number of ways to assign $n$ birth months to $k$ people  $ = \displaystyle {12 \choose n}\sum \limits_{i=0}^{n-1} {(-1)^i} {n \choose i} (n-i)^{k}$
Here $k = 25, n = 10, 11, 12$.
So for $n = 10$
$P(10) = \displaystyle \frac{{12 \choose 10}\sum \limits_{i=0}^{9} {(-1)^i} {10 \choose i} (10-i)^{25}}{12^{25}} \approx 0.302$
Similarly, $P(11) \approx 0.403, P(12) \approx 0.182$
So $P (\geq 10) \approx  0.887$
