Combination: 3 boys with 7 chairs

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SUMMARY

The problem involves seating 3 indistinguishable boys on 7 chairs arranged in a straight line, ensuring that no two boys sit next to each other. The total number of valid seating arrangements is 10, as derived from listing all possible outcomes. A generalized approach suggests using a formula where each occupied chair must have an empty chair to its right, leading to the introduction of an extra chair to facilitate the arrangement. This method can be extended to any number of boys and chairs while maintaining the same seating restriction.

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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.
 

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