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dddanny2006
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Confusion in Combinatorics methods--Manhattan--HELP!

by dddanny2006 Wed Dec 11, 2013 1:01 pm

Source-MGMAT guide

There are 7 candidates for the 4 consultant slots. If 2 out of 7 consultants refuse to be on the same team, how many different teams are possible?

Ill solve this using the slot method

Possible conditions with the constraints

7*6*5*4 = 35
4*3*2*1
When constraints are in place-- Lets assume that the 2 consultants are already selected,so what we need to do is to find the possible combinations for the other 2 consultants

5*4 = 10
2*1

So 35-10=25 possible combinations where the 2candidates are not on the same team.


My concern is why I cant use the same method in this question below

Greg, Marcia, Peter, Jan, Bobby and Cindy go to a movie and sit next to each other in 6 adjacent seats in the front row of the theater. If Marcia and Jan will not sit next to each other, in how many ways different arrangements can the 6 people sit?

So now we have 6 candidates Jan,Marcia,Greg,Peter,Bobby,Cindy

To find arrangements in which Marcia and Jan don't sit next to each other,I first find the number of arrangements in which they sit next to each other so the I can subtract those later from the total possible arrangements.

Lets assume J and M are already seated next to eachother,so for the remaining 4 spots we find the possible arrangements

4*3*2*1=24

Since order matters in which M and J sit next to each other opposed to J and M we found out earlier we multiply by 2 thus making it 48 possible arrangements.Therefore mt answer is 6!-48,but the answer is 6!-240=480 .

Why cant I use this method here??I know what the book says,but I want to know why Im wrong,I find it hard to remember too many things in P&C.
RonPurewal
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Re: Confusion in Combinatorics methods--Manhattan--HELP!

by RonPurewal Sun Dec 15, 2013 7:57 am

You're not accounting for the fact that there are several different places where M and J could be sitting next to each other.

They could be sitting next to each other in seats 1 and 2, in seats 2 and 3, in seats 3 and 4, in seats 4 and 5, or in seats 5 and 6. Your math only accounts for one of these five possibilities, so that's why the number subtracted out in the correct answer (240) is exactly five times as big as the number you've subtracted out.

By the way, I've noticed an improvement in your intuitive understanding of this material since you started posting on the forum. Good job.
dddanny2006
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Re: Confusion in Combinatorics methods--Manhattan--HELP!

by dddanny2006 Tue Dec 17, 2013 4:08 pm

Oh cool..You're videos are the best out there.
RonPurewal Wrote:You're not accounting for the fact that there are several different places where M and J could be sitting next to each other.

They could be sitting next to each other in seats 1 and 2, in seats 2 and 3, in seats 3 and 4, in seats 4 and 5, or in seats 5 and 6. Your math only accounts for one of these five possibilities, so that's why the number subtracted out in the correct answer (240) is exactly five times as big as the number you've subtracted out.

By the way, I've noticed an improvement in your intuitive understanding of this material since you started posting on the forum. Good job.
RonPurewal
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Posts: 19744
Joined: Tue Aug 14, 2007 8:23 am
 

Re: Confusion in Combinatorics methods--Manhattan--HELP!

by RonPurewal Thu Dec 19, 2013 4:19 pm

Thanks.