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Chapter 8: Problem 16

At a 12-week conference in mathematics, Sharon met seven of her friends from college. During the conference she met each friend at lunch 35 times, every pair of them 16 times, every trio eight times, every foursome four times, each set of five twice, and each set of six once, but never all seven at once. If she had lunch every day during the 84 days of the conference, did she ever have lunch alone?

Short Answer

Expert verified
Yes, Sharon had lunch alone on some days during the conference.

Step by step solution

01

Identify all combination possibilities up to 7 friends

First, find all the possibilities of friends Sharon can meet: alone, in pairs, in threes, in fours, in fives, in sixes, and in sevens. This will be accomplished using the combination theory.
02

Calculate the total lunch meetings for each combination

Next, each possibility will be multiplied by the number of corresponding meetings. For example, Sharon had lunch with each individual friend 35 times, every pair of friends 16 times, every trio 8 times, every foursome 4 times, every set of five 2 times, and every set of six once. In mathematics, a combination is represented as \( C(n,r) = \frac{n!}{r!(n-r)!}\), where \( n \) is the total number of objects or people, and \( r \) is the number of objects or people to be chosen at a time. Calculate the total lunch meetings using combinatorics.
03

Add up total lunch meetings

Add up all the lunch meetings calculated in step 2. The result should equal the total days of the conference (84 days) if Sharon didn't have lunch alone. If the total lunch meetings are less than 84, then it would signify that Sharon indeed had lunch alone on some days.

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