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?
+1223 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.
+237 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
