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rsala004 Nov17-09 06:38 PM

# of combinations
 
Lets say you have a bunch of projects to do, says project 1,2,3,4,5.

You dont have to do them all, in fact you dont have to do any of them...and the order you did them in has no effect on how they come out.

how many ways can this be done? examples, 12, 1234, 234 or no projects at all

edit:
My quick assumption to solving this problem is that you either do the project or you dont...so i guess you have 2 choices 5 times.
2*2*2*2*2 = 32, is this the solution?

Mute Nov17-09 07:26 PM

Re: # of combinations
 
I would ask: how many ways can you do 0 of the projects? 1 project? 2, 3, 4, all of them?

The total number of ways should then be (# ways 0)(# ways 1)(# ways 2)(# ways 3)(# ways 4)(# ways 5), which is more than 32.

pbandjay Nov17-09 07:51 PM

Re: # of combinations
 
This one is easy enough to just list by hand, considering the number of combinations of picking zero to five of the projects. It's 32.

The long way: Number of ways to pick 0 + Number of ways to pick 1 + ... + Number of ways to pick 5 = 1 + 5 + 10 + 10 + 5 + 1 = 32.

Or: 25 = 32.

The connection is the beautiful theorem:

[tex]\displaystyle\sum_{k=0}^n\binom{n}{k}=2^n[/tex]

Bingk Nov17-09 10:22 PM

Re: # of combinations
 
Just to be sure ... if the person decided to do project 3 first, then 5, and nothing else (i.e. 35), you're saying that would be the same as doing 5 first then 3 (i.e 53 = 35)?

If that's the case, then yes, I believe you have it correct, 32 outcomes.

But, if you mean that you can do the projects in any order, but they still count as a distinct way of doing it (i.e. 12345 and 54321 are two ways of doing it), then it would be more than 32 ...

arildno Nov18-09 02:08 AM

Re: # of combinations
 
Quote:

Quote by Bingk (Post 2447368)
Just to be sure ... if the person decided to do project 3 first, then 5, and nothing else (i.e. 35), you're saying that would be the same as doing 5 first then 3 (i.e 53 = 35)?

If that's the case, then yes, I believe you have it correct, 32 outcomes.

But, if you mean that you can do the projects in any order, but they still count as a distinct way of doing it (i.e. 12345 and 54321 are two ways of doing it), then it would be more than 32 ...

True enough!

In that case, the total number will be:
[tex]\sum_{k=0}^{n}k!\binom{n}{k}[/tex]
For n=5, this amounts to:
[tex]1*1+1*5+2*10+6*10+24*5+120*1=326[/tex]
distinct ways.

CRGreathouse Nov18-09 08:04 AM

Re: # of combinations
 
I agree, 2^5 is the answer the OP wants.

Quote:

Quote by arildno (Post 2447554)
In that case, the total number will be:
[tex]\sum_{k=0}^{n}k!\binom{n}{k}[/tex]
For n=5, this amounts to:
[tex]1*1+1*5+2*10+6*10+24*5+120*1=326[/tex]
distinct ways.

http://www.research.att.com/~njas/sequences/A000522

rsala004 Nov18-09 11:20 AM

Re: # of combinations
 
thank you


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