When it says at least with same

Question

There is a a phone support center that helps customers,which are 5 employees.How much should we do at least to speak with the same (1) employee,at least 3 times? My thought of solving it is,there are 5 employees and 3 times we try.We have repetition.My aspect of view is that i will use the type C(n,r)=n!/(n-r)!.The reason i chose it is,because i have repetition.I would give n=5 and r=3.As a result it would be 5!=(5-3)! which leads to 1*2*3*4*5 / 2!= 3*4*5=12*5= 60. Am i right?But in the end,it says with the same employee.I know i am wrong somewhere

Answer 1

This is an application of the pigeonhole principle - the clue is that we are looking for a number of times we have to repeat something before we can guarantee a particular outcome.

To solve it, you have to consider the worst case. What's the worst scenario if you call them 5 times? Probably getting each person once, since getting even one person twice would be more helpful. What happens if you call 10 times? What happens after that?

Answer 2

You are presumably being asked how many times you need to call to speak to the same person three times. If you call three times, you might speak to three different people. How many calls does it take to guarantee that you speak to the same person twice? Now how many does it take to guarantee that you speak to the same person three times?