A group of sixty people is selected at random. What is the probability that at least two of them will have the same birthday?

i am going to ignore the 29th feb

so there are 365 days in a year.

what is the prob that their birthdays are all different

$$\\P(all \;diff)=1*\frac{364}{365}*\frac{363}{365}*...... \frac{365-n+1}{365}*\frac{365-60+1}{365}\\\\
P(all \;diff)=1*\frac{364}{365}*\frac{363}{365}*...... \frac{366-n}{365}*\frac{306}{365}\\\\
P(all \;diff)=\frac{364!}{305!(365)^{59}}\\\\
P( at least 2 the same)=1-\frac{364!}{305!(365)^{59}}$$

The calc is refusing to calculate this.   I'll come back to it if time. 

Anyway it is very likely.