Combination: 3 boys with 7 chairs

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The discussion revolves around seating 3 indistinguishable boys on 7 chairs arranged in a line, ensuring that no two boys sit next to each other. The user initially lists 10 possible arrangements but seeks a generalized method for determining the seating configurations. A proposed approach involves representing occupied chairs with 'O' and empty chairs with 'X', and suggests pairing each occupied chair with an adjacent empty chair to maintain the required separation. By introducing an additional guaranteed vacant chair, the problem can be generalized for any number of boys and chairs. The conversation concludes with acknowledgment of the helpfulness of the hint provided.
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Homework Statement



Suppose there is 7 chairs arranged in a straight line, each of the 3 boys will sit randomly on one of the chair . In how many ways the boys can be seated if the 3 boys cannot sit next to each other? Assume that the boys are indistinguishable.

I listed out all the possible outcomes (which is 10), but i believe there is a generalized way to find the answer. Can anyone enlighten me?

Homework Equations





The Attempt at a Solution



Let O represent seat occupied by the boys and X is empty seat.

Possible outcomes:

XOXOXOX
XOXOXXO
XOXXOXO
XXOXOXO
OXXOXOX
OXOXXOX
OXOXOXX
OXXOXXO
OXOXXXO
OXXXOXO
 
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Michael_Light said:

Homework Statement



Suppose there is 7 chairs arranged in a straight line, each of the 3 boys will sit randomly on one of the chair . In how many ways the boys can be seated if the 3 boys cannot sit next to each other? Assume that the boys are indistinguishable.

I listed out all the possible outcomes (which is 10), but i believe there is a generalized way to find the answer. Can anyone enlighten me?

Homework Equations





The Attempt at a Solution



Let O represent seat occupied by the boys and X is empty seat.

Possible outcomes:

XOXOXOX
XOXOXXO
XOXXOXO
XXOXOXO
OXXOXOX
OXOXXOX
OXOXOXX
OXXOXXO
OXOXXXO
OXXXOXO

Using 'b' for 'boy' and 'e' for 'empty', start with bebeb and just figure out how many ways to add the two remaining 'e's.
 
For a generalized approach, suppose C chairs and B boys, same restriction. Each occupied chair, except the rightmost, must have a vacant chair on its right. To handle that exception, introduce an extra chair on the right, guaranteed vacant. So we can pair up each occupied chair with that adjacent vacant chair, making B such pairs and C+1-2B other vacant chairs. Can you proceed from there?
 
haruspex said:
For a generalized approach, suppose C chairs and B boys, same restriction. Each occupied chair, except the rightmost, must have a vacant chair on its right. To handle that exception, introduce an extra chair on the right, guaranteed vacant. So we can pair up each occupied chair with that adjacent vacant chair, making B such pairs and C+1-2B other vacant chairs. Can you proceed from there?

Got it. Your hint is very useful. Thanks.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

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