We have r people and 365 days , so total cases are (365)^r
P(at least 3 have same day bday)=1-P( no two have same day bday) --( exactly 2 have same day bday) - equation one
Case1: P( no two have same day bday)= all have different day bday
So it we have 4 people we have 4 days then and all have different bdays in 4! Ways,so here we have r people we will select r days by 365 C r * r! Ways which is 365.364....(365-(r-1))/(365)^r
Case 2: exactly 2 have same day bday :-We will select 2 out of r people have treat them as single entity and they can have bday on same day and that can be any one of 365 days,so 365 ways i.e rC2*365 ,we have 364 days remaining and r-2 people which have bday on diff day so similar to case 1 we have total case=(364)^r-2 and favourable will be selecting r-2 days out of 364 and we have 364 C r-2 * (r-2)! = 364*363....(364-(r-3) ) the last bolded term can be written as (365-(r-2))
So we get rC2* 365 * 364*363.....(365-(r-2))/(364)^r-2
Substitute both cases in equation one ,Answer will be C