# Non-Homework Combinatorics Question for a Legal Brief

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1. Apr 22, 2015

### Mark_809

1. The problem statement, all variables and given/known data

To make sure I don't get booted right off the bat, I should emphasize that this is not a homework assignment. I'm a lawyer working on a brief, and I'm trying to respond to an allegation that has dragged combinatorics into the case.

In a complaint filed in court, you are not permitted to make formulaic allegations that don't include sufficient factual allegations to properly state a claim. My problem is that a plaintiff has filed a complaint against my client (and 70+ other defendants) alleging generally that each of the 70+ defendants worked with one or more of nine different toxic substances and that they then performed one or more of eleven different activities with one or more of the nine different toxic substances. I've already referred to this as the PowerBall Approach in my brief, and a friend with a math background was able to tell me that it falls into the realm of combinatorics. But beyond that, I'm simply out of my element. I've already argued that if each defendant worked with only one substance and did only one activity, it would still describe 99 possible claims against each defendant, but that's as far as my math background can take me.

2. Relevant equations

No idea, though if my situation were simpler it would apparently be (n+m-1)!/(n-1)!*m! but it's not that simple.

3. The attempt at a solution

I tried finding an online combinatorics calculator (I was surprised they actually exist), but I think this is another level of complexity beyond that. For instance, those calculators could tell me how many ways you can get 2 substances out of nine substances (45), but it doesn't have to be 2 substances. It can be anywhere from 1 to nine. And then, it can be any one or more of eleven different activities with all of those possible combinations of substances. I want to give the judge a number and an equation to show where the number comes from to make clear how vague and all encompassing that count of the complaint actually is.

Thanks (sorry I couldn't make a better attempt at the math but this is definitely not what they taught us in law school).

2. Apr 22, 2015

### Staff: Mentor

What exactly do you want to calculate?

If you ask how many possible combinations are there in total: you could make a 9x11 matrix for each defendant, where each box corresponds to "did this activity with this substance". That gives 99 independent chemical/activity combinations, for a total of 299-1 or roughly 1030 possible ways to fill that matrix for a single defendant. The "-1" as an empty matrix would mean no substance was used for those activities.
Even with three defendants the number of options exceeds the number of particles in the known universe.

If you know that all chemicals are claimed to be used in all activities, the number of options gets much lower. You have 29-1 possible combinations of chemicals and 211-1 possible sets of activities. Still about a million options for each, and more options than atoms in the known universe for all 70 combined.

3. Apr 22, 2015

### Mark_809

I would say I'm trying to determine how many possible combinations there are for any one defendant. So Defendant A is alleged to have done one or more of eleven activities with one or more of nine different substances. What I would like to find is X in this hypothetical sentence: "By combining one or more of eleven possible activities and one or more of nine possible substances, Plaintiff's complaint presents any one of X possible allegations." And then I'd like to be able to put the mathematical equation in a footnote to show I didn't pull X out of the air.

4. Apr 22, 2015

### LCKurtz

If person X chooses some number of m drugs he has $2^m$ ways he could choose the drugs. Then if he chooses some number of n activities he has $2^n$ activities choice selections. The total number drug,activity actions for this person would be $2^m2^n=2^{m+n}=2^{20}=1048576$. That includes everything from doing 1 activity with 1 drug to doing 9 activities with 11 drugs.

5. Apr 22, 2015

### Staff: Mentor

That is the second calculation I mentioned.

You get even more if you allow things like "he did activity 1 and 3 with drug 1, activity 3, 6 and 7 with drug 4, ..."

6. Apr 22, 2015

### Mark_809

Does that also include the possibility of 0 drugs and 0 activities? (I was trying to sort of test 2^m by substituting 3 for m so it's drugs, A, B, or C - but I only came up with seven combinations: A, B, C, AB, AC, BC, or ABC - does this equation include the possibility of no drugs at all?)

7. Apr 22, 2015

### LCKurtz

I was just getting ready to edit my post when yours came in. I think the formula does account for those but it also counts for choosing 0 drugs and doing some activities or doing some drugs and 0 activities. So I think it needs to be reduced by $2^{11}+2^9-1=1535$

8. Apr 22, 2015

### Mark_809

Ah ok so would it be correct to say that if the rule is I have to have at least at least 1 drug(m) and at least 1 activities(n) the equation would be (2^9 - 1)*(2^11 - 1) = X?

9. Apr 22, 2015

### Staff: Mentor

Right, that gives about one million. Assuming you don't care which specific activity has been performed with which substance.

10. Apr 22, 2015

### LCKurtz

I agree.

11. Apr 22, 2015

### Mark_809

Exactly. So then it would work out like this...

(2^9 - 1)*(2^11 - 1) = X
(511)*(2047) = 1,046,017

Is that right?

12. Apr 22, 2015

### Mark_809

Just out of curiosity... the online combinatorics calculators used this formula to calculate how many different ways you could get any 2 out of 9 options (for instance): (n+m-1)!/(n-1)!*m!

What does the "!" mean? That formula from the calculator seems completely different from the one we just generated, which seems more straight forward.

13. Apr 22, 2015

### Mark_809

Last question (hopefully) - does our formula [(2^9 - 1)*(2^11 - 1) = X] also cover the possibility of two or more drugs combined with two or more activities?

14. Apr 22, 2015

### Staff: Mentor

! means factorial. 4! = 1*2*3*4
That is an unnecessary detour.
Sure!

15. Apr 22, 2015

### Mark_809

Thanks guys! I really appreciate it. I'm hoping this site is also good for possibly stupid questions I come up with based on my hard science fiction reading :P

Thanks again!

-Mark