Calculate # of Combinations: Projects 1-5

[rsala004]

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

You don't have to do them all, in fact you don't 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]

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]

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:

\displaystyle\sum_{k=0}^n\binom{n}{k}=2^n

 

[Bingk]

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]

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:
\sum_{k=0}^{n}k!\binom{n}{k}
For n=5, this amounts to:
1*1+1*5+2*10+6*10+24*5+120*1=326
distinct ways.

 

[CRGreathouse]

I agree, 2^5 is the answer the OP wants.

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

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

 

[rsala004]

thank you