What is the smallest number of people in a group, so that it is guaranteed that at least three of them will have their birthday in the same month?

Solution 1:

Q: How do you avoid having three people with birthdays in the same month while making your group of people as large as possible?

A: By having two in each month. That makes $24$ people. So in which month was the $25$th one born?

Solution 2:

Your question is a straightforward application of the pigeonhole principle. In its simplest form, applied to the context of your question, the pigeonhole principle states that for $m = 12$ months, if there are $n \ge 13$ people in a group, then there is guaranteed to be a month in which at least two people's birthdays occur.

This makes intuitive sense: if we had $n = 12$ people, then each person could be assigned a different birth month. But if we add in one more person, this thirteenth person would necessarily have a birth month in common with one of the other twelve.

The generalization of this idea is given in the Wikipedia article we linked to:

For natural numbers $k$ and $m$, if $n = km + 1$ objects are distributed among $m$ sets, then...at least one of the sets will contain at least $k+1$ objects.

As this applies to your original question, then, we see that $m = 12$ (the number of months), $n$ is the number of people in the group, and $k = 2$ is the maximum number of people allowed to share a birth month before satisfying the criterion of the problem.

Solution 3:

A group of twenty-four people can fail to satisfy the request (how?). What if you ask another person to join the group?

Solution 4:

12 month * 2 people + 1 maybe.

Solution 5:

The Answer is $25$, So the question is assuming that if there were $12$ people in the same room they are all born in separate months, so you would do $2\times12=24$. Now $2$ people are in each month now add one person because whichever month he is born in will allow there to be $3$ in one month.