Question

I run a book club with $n$ people, not including myself. Every day, for $365$ days, I invite five 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 five members over all $365$ days?

Answer 1

Answer:

Step-by-step explanation:

if we have n people and we want to do groups of 5, the total number of different combinations is:

$c = \frac{n!}{(n - 5)!5!}$

find the smallest n such c > 365.

let's do it by brute force:

if n = 5, we have:

$c = 5C_5 = 5$

if n = 6

$c =6C_5 =6$

if n = 7

$c = 7C_ 5= 21$

if n = 8

$c = 8C_5 = 56$

if n = 9

$C = 9!/(4!*5!) = 126$

so 9 is not enough, let's see N = 10

C = 10!/(5!*5!)  = 252

10 is not enough, let's see with 11.

C = 11!/(6!*5!) =  462

So you need at least 11 members in the club.

Then the minimum number of members such we have more than 365 combinations is 11 members

Answer 2

Answer: 11 members.

Step-by-step explanation:

When we have a number N of objects and we want to divide it into groups K objects, the number of possible combinations is:

$c = \frac{N!}{(N -K)!*K!}$

Here we have that C must be at least, 365 and K = 5, we want to find the value of N.

$365 \leq \frac{N!}{(N-5)!*5!}$

Now, you can play with different numbers of N, for example, if we take N = 9 we have:

C = 9!/(4!*5!) = 9*8*7*6/(4*3*2) = 126

so 9 is not enough, let's see N = 10

C = 10!/(5!*5!) = 10*9*8*7*6/(5*4*3*2) = 252

10 is not enough, let's see with 11.

C = 11!/(6!*5!) = 11*10*9*8*7/(5*4*3*2) = 462

So you need at least 11 members in the club.