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icedoverfire
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Combinatorics - Order of presentations 01 May 2018, 09:10
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6 people are to give presentations. Luke's presentation references something that Ruth's presentation states, so Ruth's presentation must come before Luke's, but not necessarily immediately before. How many different orders are possible?

NB: This question is from a lecture. No answer choices were given. There is an explanation but I'm just not understanding it. Help, please?

- ice

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Combinatorics - Order of presentations 01 May 2018, 09:38
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IS answer for this question is 360
that is 6!/2.
Total ways for presentation is 6! out of which in half ruth will be ahead of luke
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Hero8888
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Combinatorics - Order of presentations 01 May 2018, 10:13
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icedoverfire
6 people are to give presentations. Luke's presentation references something that Ruth's presentation states, so Ruth's presentation must come before Luke's, but not necessarily immediately before. How many different orders are possible?

NB: This question is from a lecture. No answer choices were given. There is an explanation but I'm just not understanding it. Help, please?

- ice
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Here is my approach (for guys like I, who struggle with combination problems):

When Ruth chooses to be
1st in a row (1 opion), then Luke has 5 options, and other guys 4!
2nd in a row (1 opion), then Luke has 4 options, and other guys 4!
3rd in a row (1 opion), then Luke has 3 options, and other guys 4!
4th in a row (1 opion), then Luke has 2 options, and other guys 4!
5th in a row (1 opion), then Luke has 1 options, and other guys 4!
6th - she can not because she has to be before Luke

So if you sum all these numbers (5+4+3+2+1)*4! you will get 15*4! = 360 combinations.
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icedoverfire
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Combinatorics - Order of presentations 01 May 2018, 10:31
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Hero8888
icedoverfire
6 people are to give presentations. Luke's presentation references something that Ruth's presentation states, so Ruth's presentation must come before Luke's, but not necessarily immediately before. How many different orders are possible?

NB: This question is from a lecture. No answer choices were given. There is an explanation but I'm just not understanding it. Help, please?

- ice

Here is my approach (for guys like I, who struggle with combination problems):

When Ruth chooses to be
1st in a row (1 opion), then Luke has 5 options, and other guys 4!
2nd in a row (1 opion), then Luke has 4 options, and other guys 4!
3rd in a row (1 opion), then Luke has 3 options, and other guys 4!
4th in a row (1 opion), then Luke has 2 options, and other guys 4!
5th in a row (1 opion), then Luke has 1 options, and other guys 4!
6th - she can not because she has to be before Luke

So if you sum all these numbers (5+4+3+2+1)*4! you will get 15*4! = 360 combinations.
Show more

This is exactly what I was trying to figure out. Thanks!


Sent from my iPhone using GMAT Club Forum mobile app

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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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