Find the probability that in a group of 23 people, exactly 3 people have birthdays on the same day My approach was as follows, select 3 people from 23 and then assign any one of the 365 days, then assign the remaining 20 people any of the 364 days and divide it by the total possibilities, which comes out to be the following :-
$\frac{{23\choose3}*365*^{364}P_{20}}{365^{23}}$
I would like to know what is wrong in my approach, because if you consider the number of people who have their birthday on the same day as a variable and sum the above formula from 2 to 23 it should lead to an answer close to 0.5, which is not happening. 
 A: I had a mistake in my original answer, here is a corrected version.
here is how you can calculate the probability that in a group of $23$ people exactly $3$ have the same birthday and the remaining $20$ persons all have different birthdays (so that there is a total of $21$ different birthdays).


*

*There are $23\choose{3} $ possibilities of fixing three people. 

*The probability that three persons have the same birthday is $p_1 = \frac{365}{365^3}$. 

*The probability that the remaining 20 persons have all a different birthday is $p_2 = \frac{364 \cdot \dots \cdot 345}{365^{20}}$. 


Therefore we arrive at
$$
p = {23\choose 3} \cdot p_1 \cdot p_2 = 0.007395218.
$$
I also wrote a small R simulation:
library(dplyr)

bd3 <- function(){
  x <- sample(1:365, 23, replace = T)
  n1 <- table(x) %>% as.data.frame() %>% filter(Freq==3) %>% nrow()
  n2 <- table(x) %>% as.data.frame() %>% filter(Freq==1) %>% nrow()
  return(ifelse(n1 == 1 & n2 == 20, 1 ,0))
}

set.seed(118)
replicate(100000, bd3()) %>% mean
[1] 0.0074

So the result of the simulation is close to the theoretical answer.
