Mathematics Asked by Keshav Vinayak Jha on November 30, 2020
How many people are needed to guarantee that at least two were born on the same day of the week and in the same month (perhaps in different years) ?
I couldn’t understand the wording of this question, so please explain the exact solution which is required.
Thanks!
We need to find the number of pigeonholes. Two birthdays are in the same pigeonhole if they occur in the same month on the same day. So the different pigeonholes are of the form $(text{Month, Day})$ i.e. $(text{January, Monday}),(text{January, Tuesday}),...,(text{January, Sunday}),(text{February, Monday}),...,(text{February, Sunday})...$
and so on, giving a total of $7times 12 =84$ pigeonholes. So you can take $>84$ people.
Correct answer by Shubham Johri on November 30, 2020
I will assume that the problem asks for both conditions to be satisfied (i.e, the two people must be born on both the same day of the week and in the same month). Then we have that there are $84 = 7cdot 12$ combinations of days of the week and months ("holes"), so then the number of people ("pigeons") such that pigeons must be in the same hole is $boxed{84 + 1 = 85}$
We don't have to deal with leap-years or anything like that because each month always contains more than $7$ days (and thus every day of the week).
Answered by Joshua Wang on November 30, 2020
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