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arrangements

by confuse mind » Mon Aug 20, 2012 8:43 am
What is the number of ways in which 5 books can be arranged 3 at a time?

The solution will be 5P3 * 2! or 5P3
The language of above question always confused me in the sense that the left over 2 books are also to b arranged or not?

What is the number of ways in which 6 books can be arranged 3 at a time?
6P3 * 3! or 6P3
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by Brent@GMATPrepNow » Mon Aug 20, 2012 11:12 am
confuse mind wrote:What is the number of ways in which 5 books can be arranged 3 at a time?

The solution will be 5P3 * 2! or 5P3
The language of above question always confused me in the sense that the left over 2 books are also to b arranged or not?

What is the number of ways in which 6 books can be arranged 3 at a time?
6P3 * 3! or 6P3
For your question, it looks like we don't need to worry about the remaining books (so the answers are 5P3 and 6P3).

By the way, we can answer this question without permutation notation.
We can take the task of arranging 3 books (from 5 books), and break it into stages.

Stage 1: Select a book for the first position.
This can be accomplished in 5 ways

Stage 2: Select a book for the middle position.
Since we already selected a book for stage 1, there are 4 books remaining, so this stage can be accomplished in 4 ways.

Stage 3: Select a book for the last position.
This stage can be accomplished in 3 ways.

At this point, we can apply the Fundamental Counting Principle (FCP), to see that all 3 stages can be completed in 5x4x3 ways (60 ways)

Cheers,
Brent

PS: For more information about the FCP, we have this free video: https://www.gmatprepnow.com/module/gmat-counting?id=775
Brent Hanneson - Creator of GMATPrepNow.com
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by mssharsha » Fri Aug 24, 2012 12:29 am
5 books can be arrannged taking 3 at a time in 5!/(5-3)! ways i.e 5p3=60
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by GaneshMalkar » Fri Aug 24, 2012 8:56 pm
Will make it little bit easy to understand....

Concept of Permutation
nPr = nCr * r!

Which can be said "selecting r things from n distinct things(upto this is selecting or combination) and then arranging those r things within themselves(this arrangement is Permutation)!"

so here we have a task of selecting 3 from 5 books which is nothing but 5C3....

and then arranging those three books within themselves can be done in 3! different ways

so total selection and arrangement = 5C3 * 3! = 60
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