Question

A state has a population of 36 million people. Show that there are at least 6 people in the state who were born on the same day of the year (but not necessarily in the same year) with the same three initials. (Assume that everyone’s name has three initials.)

Answer 1

Answer:

This question is based on pigeonhole principle.

Let us assume there are 365 days in a year.

As there are 26 alphabets in English, we can say that there is a total of $26^{3}$ = 17576 possible initials for each person.

Hence, there will be $17576\times365$ = 6415240 possible triple initials and date of birth for each person.

As per pigeonhole principle, there are 36,000,000 pigeons.

Therefore, we can say that there will be $\frac{36000000}{6415240}$ = 5.6 ≈ 6 persons with the same initials who were born on the same day of the same month.