# probability that no two people are born on the same day: $n$ people and $k$ days

Say we have $$n$$ people and $$k$$ days. What is the probability that NO two of the $$n$$ people are born on the same day?

This is of course assuming that $$n \leq k$$ (otherwise the answer is $$0$$ by the pigeonhole principle!). I said the answer was: $$k \choose n\cdot n! \cdot \frac{1}{k^n}$$ because in order for this to happen, n distinct birthdays must be chosen, they can be arranged in any way amongst the $$n$$ people, and there are $$k^n$$ total arrangements. Is this correct?

• Your answer appears correct to me. A good way to make sure of this would be to check for small values of $n$ and $k$ that you could count all the possible outcomes by hand and solve the probability that way and see if your formula simplifies to the same answer. – WaveX Oct 15 '18 at 18:27
• This is the well-known "birthday problem". Your answer appears to be correct. Also if $n > k$ your formula still holds, because then ${k \choose n}$ is zero by definition. – Michael Lugo Oct 15 '18 at 19:38
• Thank you both! Didn't realize it's a common problem. – 0k33 Oct 15 '18 at 19:52