We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

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

#1**+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**+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