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I run a book club with n people, not including myself. Every day, for 100 days, I invite 4 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 4 members over all 100 days?

Feb 3, 2023

### 7+0 Answers

#1
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We are looking for the smallest n such that n(n - 1)(n - 2)(n - 3) > 100.

Since 4*3*2*1 = 24 and 5*4*3*2 = 120, the smallest n that works is n = 5.

Feb 3, 2023
#2
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It is incorrect

Feb 3, 2023
#4
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n C 4 >=100

The smallest positive integer ==9

Because: 9 C 4 ==126 - unique combinations.

Guest Feb 3, 2023
#3
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For each day, we have 4 distinct members to pick from. Hence, the number of possible combinations of 4 members is given by n choose 4, or nC4. The smallest positive integer n such that nC4 is greater than or equal to 100 is n = 18. This means that with a club of at least 18 members, it's possible to pick 4 different members every day for 100 days without ever having the exact same group twice.

Feb 3, 2023
#7
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18 is the right answer.  Thank you!

Guest Feb 11, 2023
#5
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The smallest integer n that satisfies the condition is 23

This can be shown by using the Pigeonhole Principle. If there are 22 or fewer people in the club, then at least two of the groups of 4 must be identical because there are only 100 possible groupings and 100 days, which is not enough to ensure that each possible group of 4 is chosen exactly once. If there are 23 people, then there are exactly 100 possible groupings, ensuring that each group of 4 is chosen exactly once over the 100 days.

Feb 3, 2023
#6
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The smallest positive integer n so that you can avoid ever having the exact same group of 4 members over all 100 days is n = 16. This is because the total number of possible combinations of 4 members out of a group of n is n!/(n-4)! and for n = 16, this gives 4,096 combinations. Since you have 100 days, this means you can have more than 4 combinations per day and still avoid ever having the exact same group of 4 members.

Guest Feb 3, 2023