+0

0
126
2

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
+1224
+1

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
+231
0

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