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Invited guests to a party...

Problem Solving — algebra and arithmetic
Expert replies
by nysnowboard » Thu Mar 18, 2010 11:39 pm
Hey all,


This is another non-GMAT type question but understanding the solution is definitely something that could be expected of you during the GMAT (based on my limited experience with problem solving, however)

Anyways, this is what I found, and this is the solution I came up with. Since there was no solution supplied I just wanted to check my analysis and application of the permutation rules. Hope this can help someone other than myself :idea:

Problem:

You are hosting a party and have invited 5 men and 9 women. How many ways can
you seat your guests such that no two men sit together if

A.) you seat them in a line?

B.) you seat them at a round table?



My Solution:
[spoiler]
For starters, although the problem doesn't explicitly state it, I am assuming that each person is considered a distinct item.
No repetition or replacement (one person per seat and one seat per person)

So for the line seating:

14 possible seats, 9 women = 14_P_9 = 14!/5!

I analyzed the problem like this: No men can be next to each other, is the same way as saying the men must be placed in the "spaces" between the women. There are 10 of these spaces in a line and 9 in a circle.

10 spaces for the men, 5 men = 10_P_5 = 10!/5!

In total possible positions for women x men = 14!10! / 5!5!



The circular problem is similar but the spaces for the men goes from 10 to 9:

Women is same: 14!/5!

But now men is:

9 spaces, 5 men: 9_P_5 = 9!/4!

In total: 14!9! / 5!4!


I am about 80% sure of this answer but I'm a little shaky with counting so if anyone sees any glaring logic error please point it out to me. This is how we learn right? Thanks!!!


[/spoiler]
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Source: — Quantitative Reasoning |

by kstv » Fri Mar 19, 2010 6:15 pm
I guess it is not complete off GMAT menu cos if you give a search "seating in a circle " you get couple of threads.
Try this https://www.beatthegmat.com/permutation-t44229.html
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