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scoowhoop: Junior | Next Rank: 30 Posts; Posts: 28; Joined: Fri Feb 26, 2010 7:57 pm; Thanked: 1 times

by scoowhoop » Thu Apr 15, 2010 9:09 pm

When N and K are equal we use the formal (N-1)!, however what do you do when N and K do not equal?

e.g.

How many ways can 7 people sit at a circular table with only 5 seats?

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Source: Beat The GMAT — Problem Solving | Thu Apr 15, 2010 9:09 pm

liferocks: Legendary Member; Posts: 576; Joined: Sat Mar 13, 2010 8:31 pmFollowed by: 1 member; Thanked: 97 times

by liferocks » Thu Apr 15, 2010 9:20 pm

I think in this case first we select and then arrange
in your example first we select 5 people from 7 in 7C5 ways
then arrange them in (5-1)! ways

hence total ways are (7C5)4! ways

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scoowhoop: Junior | Next Rank: 30 Posts; Posts: 28; Joined: Fri Feb 26, 2010 7:57 pm; Thanked: 1 times

by scoowhoop » Thu Apr 15, 2010 10:29 pm

hence total ways are (7C5)4! ways

Are you saying (7C5)*4!, if so what does 7C5 equal? I'm not familar with this terminology. Thanks for the relpy!

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scoowhoop: Junior | Next Rank: 30 Posts; Posts: 28; Joined: Fri Feb 26, 2010 7:57 pm; Thanked: 1 times

by scoowhoop » Thu Apr 15, 2010 11:01 pm

is it.....???

7C5 = 7!/(5!*2!) = 21

therefore...

21*4! = 504

I would be helpful if someone could confirm this. thanks.

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Thu Apr 15, 2010 11:02 pm

scoowhoop wrote:
hence total ways are (7C5)4! ways
Are you saying (7C5)*4!, if so what does 7C5 equal? I'm not familar with this terminology. Thanks for the relpy!

Hi,

7C5 represents the combinations formula; you say it out loud as "seven choose 5".

We use the combinations formula when we're counting unordered subgroups. The general formula is:

nCk = n!/k!(n-k)!,

in which:

n = total number of objects; and
k = # of objects that we're selecting.

In the question you posted, we have 7 total people and we're selecting 5, so n=7 and k=5.

7C5 = 7!/5!(7-5)! = 7!/5!2! = 7*6/2*1 = 42/2 = 21

In other words, if you have 7 distinct objects, you can select 21 different groups of 5 of them.

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

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scoowhoop: Junior | Next Rank: 30 Posts; Posts: 28; Joined: Fri Feb 26, 2010 7:57 pm; Thanked: 1 times

by scoowhoop » Thu Apr 15, 2010 11:06 pm

Thanks Stuart!

So just to clarify the answer would be 504, correct?

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Fri Apr 16, 2010 9:43 am

scoowhoop wrote:Thanks Stuart!

So just to clarify the answer would be 504, correct?

Correct!

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