I run a book club with $n$ people, not including myself. Every day, for $365$ days, I invite four members in the club to review a book. What is the smallest positive integer $n$ so that I can avoid ever having the exact same group of four members over all $365$ days?

Respuesta :

Answer: 12

Step-by-step explanation:

If we have n people and we want to do groups of 4, the total number of different combinations is:

[tex]c = \frac{n!}{(n - 4)!4!}[/tex]

and we want find the smallest n such c > 365.

let's do it by brute force:

if n = 5, we have:

[tex]c = \frac{5!}{1!*4!} = 5[/tex]

if n = 6

[tex]c = \frac{6!}{2!*4!} = \frac{6*5}{2} = 15[/tex]

if n = 7

[tex]c = \frac{71}{3!*4!} = \frac{7*6*5}{3*2} = 35[/tex]

if n = 8

[tex]c = \frac{8!}{(4!*4!} = \frac{8*7*6*5}{4*3*2} = 70[/tex]

if n = 9

[tex]c = \frac{9!}{5!*4!} = \frac{9*8*7*6*5}{5*4*3*2} = 126[/tex]

if n = 10

[tex]c = \frac{10!}{6!*4!} = \frac{10*9*8*7}{4*3*2} = 210[/tex]

So you can see the patern here.

So we have the previous number multiplied by n and divided by n - 4

if n = 11

C = 210*11/(11- 4) = 330

if n = 12

C = 330*12/(12 - 4) = 495

Then the minimum number of members such we have more than 365 combinations is 12 members