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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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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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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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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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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.
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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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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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