Question

show that there are at least 7 people in a group of 39 million people, who have the same 3 initials and who were born on the same day of the year. (assume that a year has 365 days, and that every one of the 39 million people has 3 initials.)

Answer 1

Using the Pigeonhole Principle, we find that for the 39 million people, there are at least seven people who share the same three initials and birth date as there are only 6,415,240 possible combinations for initials and birth dates.

Step-by-step explanation:

We can use the Pigeonhole Principle to tackle this problem. The Pigeonhole Principle states that if you have more pigeons than pigeonholes, at least one pigeonhole must contain more than one pigeon. In this scenario, our 'pigeons' are the 39 million people, and the 'pigeonholes' are the possible combinations of initials and birth dates.

There are 26 possibilities for each initial, and since people have three initials, there are 26 x 26 x 26 combinations of initials possible which equals 17,576. Given that there are 365 days in a year, the total number of pigeonholes (initial-date combinations) is 17,576 x 365, which is 6,415,240.

With 39 million people, we see that 39,000,000 divided by 6,415,240 is approximately 6.08. This signifies that for every unique combination of three initials and a birth date, we have about six people. According to the Pigeonhole Principle, since we can only have whole people, we round up to say that there's at least one pigeonhole containing seven or more people with the same three initials and the same birth date, thus satisfying the condition proposed.