Birthday "Paradox" -- with a different perspective Background: Many people are familiar with the so-called Birthday "Paradox" that, in a room of 23 people, there is a better than 50/50 chance that two of them will share the same birthday. In its more general form for n people, the probability of no two people sharing the same birthday is $p(n) = \frac{365!}{ 365^n *(365-n)!}$. Similar calculations are used for understanding hash-space sizes, cryptographic attacks, etc.
Motivation: The reason for asking the following question is actually related to understanding a specific financial market behavior. However a variant on the "Birthday Paradox" problem fits exactly as an analogy and is likely to be of wider interest to people with different backgrounds. My question is therefore framed along the lines of the more familiar  "Birthday Paradox", as follows.
Situation: There are a total of 60 people in a room. Of these, it turns out that there are 11 (eleven) PAIRS of people who share the same birthday, and one TRIPLE (i.e. group of 3 people) who have the same birthday. The remaining 60 - 2*11 - 3 = 35 people have different birthdays. Assuming a population in which any day is equally likely for a birthday (i.e. ignore Feb 29th & possible seasonal effects) and, given the specified distribution of birthdays, the questioner would like to understand how likely (or unlikely) it is that these 60 people were really chosen at random. This question was originally posed on another site where it was left unanswered, but the questioner was advised to re-state the question in the form that now follows below.
Question: "If 60 people are chosen at random from a population in which any day is equally likely to be a person's birthday, what is the probability that there are 11 days on which exactly 2 people share a birthday, one day on which exactly 3 of them share a birthday, and no days on which 4 or more people share a birthday?"
 A: When in doubt, simulate. (I'm sure that you can actually put together a formula, but it will likely be painful to look at.) I'll work with the criterion being "at least one triple birthday" and "at least eleven double birthdays" -- changing the code below to checking for exactly so many birthdays is not hard.
n.sims <- 1e5
n.persons <- 60
counter <- 0
pb <- winProgressBar(max=n.sims)
for ( ii in 1:n.sims ) {
    setWinProgressBar(pb,ii,paste(ii,"of",n.sims))
    set.seed(ii)
    birthdays <- sample(x=365,size=n.persons,replace=TRUE)
    birthday.table <- table(birthdays)
    if (    sum(birthday.table>=4) == 0  &
                sum(birthday.table==3) >= 1  &
                sum(birthday.table==2) >= 11 ) counter <- counter+1
}
close(pb)
counter/n.sims

Out of 100,000 simulations, I get six hits, for a $p$-value of $p=0.00006$. If you twiddle the set.seed() command, e.g., to set.seed(2*ii), you will get slightly different results (in this case $p=0.00004$), which serves as a sort of sensitivity analysis.
A: Assuming you aim for the case that exactly (so not at least) 11 double birthdays and 1 triple birthdays occur:
$$\begin{align}p(60,n_2=11,n_3=1) &= \frac{\text{possibilities with 11 double birthdays and 1 triple birthdays}}{\text{all possibilities}}\\ 
&= \frac{\frac{60!}{35!22!3!} 21!! \,\cdot \,365 \cdot 364 \cdot \, ...  \, \cdot (365-n+1+1\cdot11+2\cdot1)}{365^{60}}\end{align}$$
which my calculation in R 
factorial(60)/(factorial(35)*factorial(22)*factorial(3))*
pracma::factorial2(21)*
cumprod(319:365)[47]/365^60

approximates as $3.64 * 10^{-5}$
Explanation of the terms in the calculation


*

*$365^{60}$ is the number of ways to select random 60 birthdays among equally probable 365 days.

*$365 \cdot 364 \cdot \, ...  \, \cdot (365-n+1+1\cdot11+2\cdot1)$ is the number of ways to select random 60-11-2 unique days among equally probable 365 days.

*$\frac{60!}{35!22!3!}$ is the number of unique ways to partition 60 people in groups of 35, 22, and 3 (the numbers of people with single birthdays, double birthdays, and triple birthdays)

*$21!! = 21 \cdot 19 \cdot ... \cdot 3 \cdot 1$ is the number of ways that we can partition the 22 people in the double birthdays group into pairs.

