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scott.yin
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General Help on combinatorics

by scott.yin Sun Aug 14, 2011 5:02 pm

When using factorials to solve combinatorics, can we formulate a rule for determining what is "indistinguishable", I have difficulty seeing why one some questions you need to divide by "constraint" factorials and not on others.
katemikheeva
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Re: General Help on combinatorics

by katemikheeva Tue Aug 16, 2011 10:42 am

It goes back to avoiding double counting (e.g., order matters vs doesn't matter)
scott.yin
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Re: General Help on combinatorics

by scott.yin Tue Aug 16, 2011 2:31 pm

ok so given the basic example of 4 available seats on a plane for 7 people. If order did not matter, we sould compute the number of different possibilities as 7!/4!3! ? And if order did it would be 7!/4!?
jnelson0612
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Re: General Help on combinatorics

by jnelson0612 Sun Sep 11, 2011 11:13 pm

scott.yin Wrote:ok so given the basic example of 4 available seats on a plane for 7 people. If order did not matter, we sould compute the number of different possibilities as 7!/4!3! ? And if order did it would be 7!/4!?


If order does not matter, we use Pool!/In!Out!. Thus, if we are selecting four people out of seven to sit in the seats, we do indeed have 7!/4!3!, with four people chosen and three people not chosen.

If order matters, we use a slightly different formula: Pool!/Place!Place!Place!Place!Out!. In this case, one person can sit in seat A, one person can sit in seat B, one can sit in seat C, and so on, and three will be left out.

Thus, our formula here is 7!/1!1!1!1!3!, which simplifies to 7!/3!.

Hope this helps!
Jamie Nelson
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ldesegur
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Re: General Help on combinatorics

by ldesegur Mon Sep 12, 2011 12:52 am

I got stuck on the same issue.

I resolved by thinking that when order doesn't matter I need to get rid of the "others" by dividing the result by its factorial count (the "Out!").

Thanks for the explanation of 1!... It makes things much clear, and I wished it was in the book.

Many times the book refers to combinations formulas in the exercises. The formulas are NOT easy to find back in the main text. I wish they were clearly delimited in the text.

On the arrangement with constraint, should I assume that I need to multiply the result computed with a group by the factorial of the number of elements in the group (for JM: it's 2! = 2)?
tim
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Re: General Help on combinatorics

by tim Tue Oct 04, 2011 12:47 am

Thanks Jamie, awesome explanation. Using those 1’s as placeholders is a really helpful tool that will help keep you from even having to worry about whether order matters or what you should or should not put in the denominator..

ldesegur, you’re going to have to be more specific about your question. It’s too vague for me to answer..
Tim Sanders
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