Choosing Hotels for Meetings: Combination Rule Solutions

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Homework Help Overview

The problem involves selecting hotels for two meetings from a set of six options, with variations on whether the same hotel can be chosen for both meetings. The subject area relates to combinatorial counting principles.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of tree diagrams and counting rules, questioning the applicability of permutations and combinations for the scenarios presented. There is a focus on the importance of order in the selection process.

Discussion Status

Some participants have provided insights into the reasoning behind the counting methods, suggesting that the approach to part a) is simpler than initially thought. Others are exploring the implications of the order of selection and how it affects the total number of combinations.

Contextual Notes

There is an ongoing discussion about the assumptions regarding the interchangeability of the meetings and the implications for the counting methods used. Participants are considering the constraints of the problem as they relate to the rules of combinations and permutations.

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Homework Statement

An accounting professional association is considering six hotels as possible sites for their next two meetings in 2001 and 2002. In how many ways can the association select the hotels...
a) if the two meetings may be held at the same hotel?
b)if the two meetings may not be held at the same hotel?

Homework Equations


The Attempt at a Solution


I found the answers by drawing out a tree... But i was wondering if there was a mathematical way using the counting rules i.e permutation/combinations rules etc. I was trying for a while but couldn't find one
 
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maiad said:

Homework Statement

An accounting professional association is considering six hotels as possible sites for their next two meetings in 2001 and 2002. In how many ways can the association select the hotels...
a) if the two meetings may be held at the same hotel?
b)if the two meetings may not be held at the same hotel?

Homework Equations


The Attempt at a Solution


I found the answers by drawing out a tree... But i was wondering if there was a mathematical way using the counting rules i.e permutation/combinations rules etc. I was trying for a while but couldn't find one

For part a), there are six hotels and only two meetings to consider. So you only have to concern yourself with re-arranging the possibilities of the six hotels.

##{n \choose k}##

Does this look familiar to you? How about part b)?
 
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Actually combinations will not work for a as they exclude the possibility of the hotels being the same in both years AND, more importantly, they imply the order doesn't matter while actually it is important here because the two years are not interchangeable (e.g. selecting hotel 1 then 2 is a different situation from selecting hotel 2 then 1).

The answer is actually simpler: there are 6 possible choices in year 1 and for each of those there are 6 choices in year two. How many does that give you in total? (If you look back at your tree you may see that this is basically the same calculation except without drawing the tree).

You can apply the same approach to b).
 
OH okay i get it. Ty for the explanation guys ^^
 
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