Question 3: In les Miserables Monsier, Thenardier, and Madame Thenardier are married and during act two, Cosette and Marius beco
me married. If Thenardiers must sit next to one another and Cosette and Marius must sit next to one another, but there are no restrictions on how others can sit, how many different ways can you sit 13 people around a circular table?
Note: this is a single number (not an ordered triple or a collection of three different numbers) roughly equal to about 14.5 million if you round to the nearest hundred thousand.
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Explanation:
There are 13 people. Let's call them person A, person B, person C, ... all the way up to person M. The first four people are given who we'll call A through D. The rest (E through M) aren't really important since they aren't named.
A = Monsier Thenardier B = Madame Thenardier C = Cosette D = Marius Peron's E through M = remaining 9 people
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A and B must stick together. Because of this, we can consider "AB" as one "person". So we go from 13 people to 13-2+1 = 12 "people".
Likewise, C and D must stick together. We can consider "CD" as one "person". So we go from 12 "people" to 12-2+1 = 11 "people"
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The question is now: how many ways can we arrange these 11 "people" around a circular table? The answer is (n-1)! ways where n = 11 in this case
So, (n-1)! = (11-1)! = 10! = 10*9*8*7*6*5*4*3*2*1 = 3,628,800
We're almost at the answer. We need to do two adjustments.
First off, for any single permutation, there are two ways to arrange "AB". The first is "AB" itself and the second is the reverse of that "BA". So we will multiply 3,628,800 by 2 to get 2*3,628,800 = 7,257,600
Using similar logic for "CD", we double 7,257,600 to get 2*7,257,600 = 14,515,200
The answer is number 3 because it steadily increases or goes up so it's a straight line it doesn't go down or it doesn't decrease the number of guests!
