Arranging Deputies on a Bench with Restrictions

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Homework Help Overview

The problem involves arranging ten deputies on a bench with specific restrictions regarding their placement. The Russian and American deputies do not want to sit next to each other, while the English and French deputies prefer to sit together. The discussion explores combinatorial approaches to calculate the arrangements under these conditions.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss various methods for counting arrangements, including treating certain deputies as indistinguishable and using combinatorial formulas. Some suggest alternative approaches, such as counting all permutations and then subtracting cases where the Russian and American deputies are adjacent.

Discussion Status

Several participants have offered different perspectives on the problem, with some expressing confidence in their approaches while others seek clarification on specific formulas and methods. There is an ongoing exploration of the problem without a clear consensus on the best method.

Contextual Notes

Participants are navigating the complexities of combinatorial arrangements and the implications of the deputies' preferences on their placement. There is mention of potential confusion regarding the application of certain combinatorial formulas.

quasar987
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Ten deputies must be aligned on a bench. How many ways of doing that is there, provided that the russian deputy and the american deputy do not want to be placed side to side, and that the english and french ones want to be side to side.

Since the english and the french deputies want to be side to side, let's consider them as 1 deputy for the time being (more preceisely, for the portion of the solution that treats the deputies as indistinguishable from one another).

Let's denote x1 the number of deputy placed on the russian's left, x_2 the number of deputy placed in btw them and x3 the number of deputy to the american's right. x1 and x3 can be zero but x2 must be at least 1. Of course x1+x2+x3 must be 7. The number of non-negative integer solutions to x1+x2+x3=7 under the constraint x2[itex]\geq[/itex]1, is the same as the number of positive integer solution to (x1-1)+x2+(x3-1)=7 <==> x1+x2+x3=9 Based on the fact that the number of positive integer solutions to [itex]x_1+...x_r=n[/itex] is [itex]\binom{n-1}{r-1}[/itex], the number of ways to place the deputies (indistinguisably) around the american and the russian is thus [itex]\binom{9-1}{3-1}=\binom{8}{2}[/itex].

Taking into account all the possible permutation of the 7 deputy amongs them, we get [itex]\binom{8}{2}7![/itex]. Taking into account the possible permutation of the russian and the american, we throw a 2! into the mix. And finally, in order to take under consideration the possible permutations of the ensligh and french deputy we throw one last 2! into the mix and the glorious final answer is

[tex]\binom{8}{2}7!2!^2[/tex] :approve: At first I had doubt about my solution, but it looks pretty solid now. Looks good?
 
Last edited:
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That looks fine. Another appraoch would be to count all permutations not caring where the russian and american are, then subtract the number where they are side by side.
 
Looks good, though that "number of positive solutions" formula threw me (the formula which I find easier to remember is for the number of non-negative solutions). There is a simpler way to do it:

1. Place everybody but the Russian. 8! * 2
2. Place the Russian. 7 * 8! * 2
 
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Wow neat!

shmoe, I tought about using that method at first, but could not see what an expression for the number where they are side by side would be.
 
quasar987 said:
shmoe, I tought about using that method at first, but could not see what an expression for the number where they are side by side would be.

I'm sure you can... you were able to deal with the English and French deputies getting cozy, the Russian and American are no different.
 

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