Calculate # of Combinations: Projects 1-5

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Discussion Overview

The discussion revolves around calculating the number of ways to complete a set of projects (1-5), considering various conditions such as whether projects must be completed in a specific order or not. The scope includes combinatorial reasoning and mathematical exploration.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests that since each project can either be done or not, the total combinations would be calculated as 2 choices for each of the 5 projects, leading to 32 combinations.
  • Another participant questions the initial assumption and proposes that the total number of combinations should account for the number of ways to do 0, 1, 2, 3, 4, or all 5 projects, suggesting that the total would exceed 32.
  • A different participant confirms that the total combinations can be calculated by listing all combinations, arriving at 32 through a summation of binomial coefficients.
  • Some participants clarify that if the order of projects matters, the total would be greater than 32, leading to a formula involving factorials and binomial coefficients to calculate distinct arrangements.
  • One participant agrees with the initial calculation of 2^5 but also presents a more complex formula for counting distinct arrangements when order is considered, resulting in 326 distinct ways.

Areas of Agreement / Disagreement

Participants express differing views on whether the order of projects affects the total count of combinations. While some agree on the 32 combinations under the assumption of no order, others argue that including order leads to a significantly higher total, indicating unresolved disagreement on the interpretation of the problem.

Contextual Notes

There are multiple interpretations of how to count the combinations based on whether the order of projects is considered significant or not, leading to different mathematical approaches and results.

rsala004
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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?
 
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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.
 
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]
 
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 ...
 
Bingk said:
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.
 
I agree, 2^5 is the answer the OP wants.

arildno said:
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
 
Last edited by a moderator:
thank you
 

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