This is a discrete math (combinatorial arguments) problem. Please explain clearly, no cursive writing. Show that among a group of 621 people, there are at least 21 who are born on the same day of the month (e.g., the 21st or the 12th, etc.). Is the same fact true if there are only 620 people? Make sure to clearly indicate the use of the Pigeonhole Principle.

Name: This is a discrete math (combinatorial arguments) problem. Please expl
Uploaded: 2024-03-20T07:07:37.000Z
Channel: Bartleby
Description: This is a discrete math (combinatorial arguments) problem. Please explain clearly, no cursive writing. Show that among a group of 621 people, there are at least 21 who are born on the same day of the month (e.g., the 21st or the 12th, etc.). Is the s

Given 620 people was born in the same month .without loss of generality , assume that the month had 30 days .We shall show it by contridiction assume that , there were less than equal to 20 birthatmostin each day of that month .Exceeding that , we have the result. In thirty days Total birth was≤30x20=600But the group of people had 621 people , another 21 people had toborn in some day of that month .Note that pigeon hole principal was applied hereSo , ∃ some day x in which more than twenty people born .So , there are atleast 21 who are born on the same day If there were 620 people , the result remains the same . This followsfrom the argument we had .Note that in this case , we would have 620-600=20 extra people .So , there would have some day in which more than 20 people born .Hence , we are done .