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

 Feb 1, 2022
 #1
avatar+1224 
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We want to find the smallest n such that \(\binom{n}{3} > 365.\)

\(\frac{(n)(n-1)(n-2)}{6} > 365\)

\((n)(n^2-3n+2) > 2190\)

\(n^3 - 3n^2 + 2n - 2190 > 0\)

 

n has to be 15 or greater, so the smallest positive integer n that works is 15.

 Feb 1, 2022
 #2
avatar+231 
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We need to find n such that choosing 3 people from that where order doesn't matter and the number of ways is greater than 365. We use the formula for combination:

ncr = (n!)/r!(n-r)! In this case we will --> (n!)/3!(n-3)!

                          choose 3 people 

The number of ways for this must be greater than 365:

(n!)/3!(n-3)! ≥ 364 --> (n(n-1)(n-2)(n-3)!)/6(n-3)! ≥ 365

                                    n(n-1)(n-2) ≥ 2190
Use trial and error. Use values starting from n = 11:

n n(n-1)(n-2)  
11 11(10)(9) 990
12 12(11)(10) 1320
13 13(14)(15) 1716
14 14(13)(12) 2184
15 15(14)(13) 2730


For n = 15, the number of ways of choosing 3 random people became larger than 2190.

n = 15

 Feb 2, 2022

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