+0

# Difficult Problem...

0
192
3
+394

I have \(n\) friends. Every night of the 365-day year I invite 4 of them to dinner. What is the smallest \(n\) could be such that it is still possible for me to make these invitations without ever inviting the same group of four friends?

Mr.Owl  Oct 24, 2017
Sort:

#1
+333
+1

The minimum is 7. 6 choose 4 is 360, which is a bit too low, and 8 choose 4 is 1680, but 7 choose 4 is 840, which is the minimum amount of friends you need.

But seriously, why do you want the minimum amount of friends? Don't you want to have the most friends possible?

Mathhemathh  Oct 25, 2017
#2
+85821
+3

Here's an graphical way of figuring this out

C (n , 4)   =  n! /  ( [ n - 4] !  * 4! )   .....so....we require that.......

C ( n , 4 )  ≥  365

n! /  ( [ n - 4] !  * 4! )  ≥  365

n! / [ n - 4]!  ≥   4! * 365

n! / [ n - 4]!  ≥  8760

n * (n - 1) (n - 2) (n - 3)  ≥  8760

Look  at the graph here :

https://www.desmos.com/calculator/j9rafrlbpn

Note that  the minimum n that makes this true ≈  11.3

So.....we need 12 friends

Proof  C(11, 4)  = 330    too few groups of 4

But  C( 12. 4)  = 495     .......we even have a few "leftover" groups of 4  !!!!!!

CPhill  Oct 25, 2017
#3
+92225
+1

Another one for me to study.

Thanks Chris :)

Melody  Oct 26, 2017

### 23 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details