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# An advanced Combination/Permutation question

Posted on 2003-12-07
Medium Priority
2,002 Views
Last Modified: 2008-02-01
I have 20 alphas, 20 betas and 10 gammas in alphabetical order. How many ways (in terms of outcomes not possibilities) can I then compile a second list of  alphas, betas & gammas so that 50% of the second list match the orginal list order?

ie, an abbreviated example showing an acceptable outcome

alpha     beta
..
alpha     alpha
beta       beta
..
beta       gamma
gamma   alpha
..
gamma   gamma

see http://oldlook.experts-exchange.com/Applications/MS_Office/Excel/Q_20815663.html for the background to my query

Cheers

Dave
0
Question by:Dave Brett
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25 Comments

LVL 31

Expert Comment

ID: 9894656
Have you been abled to form a 50% matching? I just quickly did a counting argument on a scrap of paper that showed it is not possible.
0

LVL 50

Author Comment

ID: 9894674
LOL - I didnt actually check :)

Can you please check out my link to the Excel question, I tried to simplify the criteria here to make it easier without actually checking if it worked

How about we make it 20 alphas,20 betas & 20 gammas. Presumably that should form a valid 50% matching l okay

Cheers
Dave
0

LVL 31

Expert Comment

ID: 9894790
Yes it is possible with 20,20 and 20 with the obvious 10 of each matching/not matching. This is an interesting problem that in the general case looks very complex but I think there may be a clever solution (possibly not though). I will look tomorrow as it way way past bed time here. GfW
0

LVL 84

Expert Comment

ID: 9895610
I count
20!*20!*10!/(13!*7!*0! * 0!*11!*9!  *  7!*2!*1!)
+
20!*20!*10!/(11!*9!*0! * 2!*11!*7! * 7!*0!*3!)
+
20!*20!*10!/(10!*10!*0!  *  5!*10!*5!  *  5!*0!*5!)
+
20!*20!*10!/(9!*11!*0!  *  8!*9!*3!  *  3!*0!*7!)
+
20!*20!*10!/(8!*12!*0!  *  11!*8!*1!  *  1!*0!*9!)
0

LVL 50

Author Comment

ID: 9895628
Hi GwynforWeb,

Yes, this is more complex that it first seems as normally a combination calc pulls out "different" specimens, ie horse placing etc. This one is a bit trickier.

Hi ozo

Can you please provide a detailed explanation for this

Cheers

Dave

0

LVL 84

Expert Comment

ID: 9895659
And for the 20 20 20 case I count
20!20!20! /
(20!*0!*0!*
0!*5!*15!*
0!*15!*5!)
+ 20!20!20! /
(18!*2!*0!*
0!*6!*14!*
2!*12!*6!)
+ 20!20!20! /
(16!*4!*0!*
0!*7!*13!*
4!*9!*7!)
+ 20!20!20! /
(14!*6!*0!*
0!*8!*12!*
6!*6!*8!)
+ 20!20!20! /
(12!*8!*0!*
0!*9!*11!*
8!*3!*9!)
+ 20!20!20! /
(10!*10!*0!*
0!*10!*10!*
10!*0!*10!)
+ 20!20!20! /
(9!*11!*0!*
3!*9!*8!*
8!*0!*12!)
+ 20!20!20! /
(8!*12!*0!*
6!*8!*6!*
6!*0!*14!)
+ 20!20!20! /
(7!*13!*0!*
9!*7!*4!*
4!*0!*16!)
+ 20!20!20! /
(6!*14!*0!*
12!*6!*2!*
2!*0!*18!)
+ 20!20!20! /
(5!*15!*0!*
15!*5!*0!*
0!*0!*20!)
0

LVL 84

Expert Comment

ID: 9895679
Wait a minute, that can't be right
0

LVL 84

Expert Comment

ID: 9899677
This should be correct now:
20!20!10! /
(8! 2! 10!
3! 17! 0!
9! 1! 0!)
+ 20!20!10! /
(8! 3! 9!
2! 17! 1!
10! 0! 0!)
+ 20!20!10! /
(8! 3! 9!
4! 16! 0!
8! 1! 1!)
+ 20!20!10! /
(8! 4! 8!
3! 16! 1!
9! 0! 1!)
+ 20!20!10! /
(8! 4! 8!
5! 15! 0!
7! 1! 2!)
+ 20!20!10! /
(8! 5! 7!
4! 15! 1!
8! 0! 2!)
+ 20!20!10! /
(8! 5! 7!
6! 14! 0!
6! 1! 3!)
+ 20!20!10! /
(8! 6! 6!
5! 14! 1!
7! 0! 3!)
+ 20!20!10! /
(8! 6! 6!
7! 13! 0!
5! 1! 4!)
+ 20!20!10! /
(8! 7! 5!
6! 13! 1!
6! 0! 4!)
+ 20!20!10! /
(8! 7! 5!
8! 12! 0!
4! 1! 5!)
+ 20!20!10! /
(8! 8! 4!
7! 12! 1!
5! 0! 5!)
+ 20!20!10! /
(8! 8! 4!
9! 11! 0!
3! 1! 6!)
+ 20!20!10! /
(8! 9! 3!
8! 11! 1!
4! 0! 6!)
+ 20!20!10! /
(8! 9! 3!
10! 10! 0!
2! 1! 7!)
+ 20!20!10! /
(8! 10! 2!
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3! 0! 7!)
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(8! 10! 2!
11! 9! 0!
1! 1! 8!)
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(8! 11! 1!
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(8! 11! 1!
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0! 1! 9!)
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(8! 12! 0!
11! 8! 1!
1! 0! 9!)
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1! 16! 3!
10! 0! 0!)
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1! 5! 4!)
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(10! 5! 5!
2! 14! 4!
8! 1! 1!)
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4! 13! 3!
6! 2! 2!)
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(10! 5! 5!
6! 12! 2!
4! 3! 3!)
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(10! 5! 5!
8! 11! 1!
2! 4! 4!)
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(10! 5! 5!
10! 10! 0!
0! 5! 5!)
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(10! 6! 4!
1! 14! 5!
9! 0! 1!)
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(10! 6! 4!
3! 13! 4!
7! 1! 2!)
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(10! 6! 4!
5! 12! 3!
5! 2! 3!)
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(10! 6! 4!
7! 11! 2!
3! 3! 4!)
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(10! 6! 4!
9! 10! 1!
1! 4! 5!)
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(10! 7! 3!
2! 13! 5!
8! 0! 2!)
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(10! 7! 3!
4! 12! 4!
6! 1! 3!)
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(10! 7! 3!
6! 11! 3!
4! 2! 4!)
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(10! 7! 3!
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2! 3! 5!)
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(10! 8! 2!
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(10! 8! 2!
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3! 2! 5!)
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(10! 8! 2!
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1! 3! 6!)
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(10! 9! 1!
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3! 14! 3!
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4! 5! 1!)
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(11! 2! 7!
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2! 6! 2!)
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0! 7! 3!)
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2! 14! 4!
7! 3! 0!)
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3! 5! 2!)
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1! 6! 3!)
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1! 14! 5!
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2! 5! 3!)
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1! 5! 4!)
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8! 1! 1!)
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0! 5! 5!)
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(13! 6! 1!
3! 10! 7!
4! 4! 2!)
+ 20!20!10! /
(13! 6! 1!
5! 9! 6!
2! 5! 3!)
+ 20!20!10! /
(13! 6! 1!
7! 8! 5!
0! 6! 4!)
+ 20!20!10! /
(13! 7! 0!
0! 11! 9!
7! 2! 1!)
+ 20!20!10! /
(13! 7! 0!
2! 10! 8!
5! 3! 2!)
+ 20!20!10! /
(13! 7! 0!
4! 9! 7!
3! 4! 3!)
+ 20!20!10! /
(13! 7! 0!
6! 8! 6!
1! 5! 4!)
+ 20!20!10! /
(14! 0! 6!
5! 11! 4!
1! 9! 0!)
+ 20!20!10! /
(14! 1! 5!
4! 11! 5!
2! 8! 0!)
+ 20!20!10! /
(14! 1! 5!
6! 10! 4!
0! 9! 1!)
+ 20!20!10! /
(14! 2! 4!
3! 11! 6!
3! 7! 0!)
+ 20!20!10! /
(14! 2! 4!
5! 10! 5!
1! 8! 1!)
+ 20!20!10! /
(14! 3! 3!
2! 11! 7!
4! 6! 0!)
+ 20!20!10! /
(14! 3! 3!
4! 10! 6!
2! 7! 1!)
+ 20!20!10! /
(14! 3! 3!
6! 9! 5!
0! 8! 2!)
+ 20!20!10! /
(14! 4! 2!
1! 11! 8!
5! 5! 0!)
+ 20!20!10! /
(14! 4! 2!
3! 10! 7!
3! 6! 1!)
+ 20!20!10! /
(14! 4! 2!
5! 9! 6!
1! 7! 2!)
+ 20!20!10! /
(14! 5! 1!
0! 11! 9!
6! 4! 0!)
+ 20!20!10! /
(14! 5! 1!
2! 10! 8!
4! 5! 1!)
+ 20!20!10! /
(14! 5! 1!
4! 9! 7!
2! 6! 2!)
+ 20!20!10! /
(14! 5! 1!
6! 8! 6!
0! 7! 3!)
+ 20!20!10! /
(14! 6! 0!
1! 10! 9!
5! 4! 1!)
+ 20!20!10! /
(14! 6! 0!
3! 9! 8!
3! 5! 2!)
+ 20!20!10! /
(14! 6! 0!
5! 8! 7!
1! 6! 3!)
+ 20!20!10! /
(15! 0! 5!
5! 10! 5!
0! 10! 0!)
+ 20!20!10! /
(15! 1! 4!
4! 10! 6!
1! 9! 0!)
+ 20!20!10! /
(15! 2! 3!
3! 10! 7!
2! 8! 0!)
+ 20!20!10! /
(15! 2! 3!
5! 9! 6!
0! 9! 1!)
+ 20!20!10! /
(15! 3! 2!
2! 10! 8!
3! 7! 0!)
+ 20!20!10! /
(15! 3! 2!
4! 9! 7!
1! 8! 1!)
+ 20!20!10! /
(15! 4! 1!
1! 10! 9!
4! 6! 0!)
+ 20!20!10! /
(15! 4! 1!
3! 9! 8!
2! 7! 1!)
+ 20!20!10! /
(15! 4! 1!
5! 8! 7!
0! 8! 2!)
+ 20!20!10! /
(15! 5! 0!
0! 10! 10!
5! 5! 0!)
+ 20!20!10! /
(15! 5! 0!
2! 9! 9!
3! 6! 1!)
+ 20!20!10! /
(15! 5! 0!
4! 8! 8!
1! 7! 2!)
+ 20!20!10! /
(16! 1! 3!
4! 9! 7!
0! 10! 0!)
+ 20!20!10! /
(16! 2! 2!
3! 9! 8!
1! 9! 0!)
+ 20!20!10! /
(16! 3! 1!
2! 9! 9!
2! 8! 0!)
+ 20!20!10! /
(16! 3! 1!
4! 8! 8!
0! 9! 1!)
+ 20!20!10! /
(16! 4! 0!
1! 9! 10!
3! 7! 0!)
+ 20!20!10! /
(16! 4! 0!
3! 8! 9!
1! 8! 1!)
+ 20!20!10! /
(17! 2! 1!
3! 8! 9!
0! 10! 0!)
+ 20!20!10! /
(17! 3! 0!
2! 8! 10!
1! 9! 0!)

I don't see a simple method of enumerating these other than brute force
0

LVL 84

Expert Comment

ID: 9902468
The formula I get for the 20 20 20 case is  a bit long to post, but the final number I get is 6129796750811232007318948
The total count for the 20 20 10 case was 215331640837573989718704
I could post the script I used to enumerate these if GwynforWeb doesn't come up with a more clever solution.
0

LVL 1

Expert Comment

ID: 9913872
Hi  brettdj,
I have read both your question and the link, I'm still not quite clear on the question.

When you say 50% of the second list matches the first list. Do you mean that at exactly 1/2 of the positions the two lists should match? i.e. the follow would be acceptable.
1    2
A   A
B   A
G   G
A   B

Does the second list should use exactly the same number of alpha, beta and gamma as the first one?
0

LVL 1

Expert Comment

ID: 9931065
Dose the formula n = 50, k = 25
n!
--------- 2^k
k! (n-k)!

work?
0

LVL 50

Author Comment

ID: 9932840
Hi ozo,

My eyes are spinning  :)

Can you please expand on your logic

Hi skp23,

Yes that would be a match

The problem with this question is that some ouctomes are duplicates, so I'd like to know how many discrete outcomes are possible rather than how many possibilities are possible. Order is important but some permutations are identical, ie

1    2     3      4      5
A    G1    G2   G1    G2
G    G2    G1   G2    G1
A    A1    A1   A2    A2
G    A2    A2   A1    A1

For example, I've shown four of the possible permuations of rearranging two gammas and two alphas so that 50% of the results correspond to group 1. But all four permuataions are an identcial outcome.

Whereas

G
A
A
G

would be another discrete outcome that satisfies the constraints

Make sense?

Cheers
Dave
0

LVL 84

Accepted Solution

ozo earned 2000 total points
ID: 9932968
The factorial coefficients in the denominator represent all the ways to fill a 3Ã—3 matrix with nonnegative integers such that the sum of the first row and column is 20, the sum of the second row and column is 20, the sum of the third row and column is 10, and the sum of the diagonal is 25
0

LVL 50

Author Comment

ID: 9936003
Can you please repeat this method on a smaller sample which I can actually test to check that it matches what I want

How about on 4 alphas, 4 betas and 2 gammas with the same 50% criteria

Sorry for stringing this out but I'm keen to understand it

Cheers

Dave
0

LVL 84

Expert Comment

ID: 9936215
4!4!2! /
(2! 0! 2!
1! 3! 0!
1! 1! 0!)
+ 4!4!2! /
(2! 1! 1!
0! 3! 1!
2! 0! 0!)
+ 4!4!2! /
(2! 1! 1!
2! 2! 0!
0! 1! 1!)
+ 4!4!2! /
(2! 2! 0!
1! 2! 1!
1! 0! 1!)
+ 4!4!2! /
(3! 0! 1!
1! 2! 1!
0! 2! 0!)
+ 4!4!2! /
(3! 1! 0!
0! 2! 2!
1! 1! 0!)
=2304
0

LVL 84

Expert Comment

ID: 9936345
Wait, that's wrong again
0

LVL 84

Expert Comment

ID: 9936421
4!4!2! /
(2! 0! 2!
1! 3! 0!
1! 1! 0!)
+ 4!4!2! /
(2! 1! 1!
0! 3! 1!
2! 0! 0!)
+ 4!4!2! /
(2! 1! 1!
2! 2! 0!
0! 1! 1!)
+ 4!4!2! /
(2! 2! 0!
1! 2! 1!
1! 0! 1!)
+ 4!4!2! /
(3! 0! 1!
1! 2! 1!
0! 2! 0!)
+ 4!4!2! /
(3! 1! 0!
0! 2! 2!
1! 1! 0!)
=480
0

LVL 50

Author Comment

ID: 9936486
Ok, I'll go away and crunch it in an excel spreadsheet  :)

Can you give me any links to look at to explain this technique?

Cheers

Dave

0

LVL 50

Author Comment

ID: 9937175
Bugger it - that still a long one to test as I'm not sure I've generated all the discrete outcomes. So....

I can see the pattern in what you doing but would like a little more understang, a link to your mathematical techique will do fine and I'll then accept the answer. I don't want to string you out too long on this

If anyone else then  posts another method or finds an error then I'll open a new question

Thanks for your hard work on this

Cheers

Dave
0

LVL 1

Expert Comment

ID: 10010970
Let you have x alpha mismatches (and 20-x matches), y beta mismatches (and 20-y matches), and z gamma mismatches (and 10-z matches). For x,y,z there are two constraints:
1) Since you want 50% matches, x+y+z=(20+20+10)/2=25
2) "Triangle" ineqalities x<=y+z, y<=x+z, z<=x+y; otherwise you won't be able to produce mismatches.

Let we have t alpha-beta mismatches (i.e. t alphas are replaced with betas in the second list). Then we have x-t alpha-gamma mismatches, y-t gamma-beta mismatches, z-y+t gamma-alpha mismatches, x-z+y-t beta-alpha mismatches, and z+t-x beta-gamma mismatches.
Number of ways to arrange them on x alpha positions, y beta positions, and z gamma positions is
C(x,t)*C(y,z+t-x)*C(z,y-t),
where C(n,m) is ``n choose m'' aka binomial coefficient.

And finally we need to arrange mismatch positions in the second list - there are C(20,x)*C(20,y)*C(10,z) ways.

So, number of the second lists (mis)matching exactly 50% of the original list is
Sum(C(x,t)*C(y,z+t-x)*C(z,y-t)*C(20,x)*C(20,y)*C(10,z)), where sum is taken over all suitable x,y,z,t.

In mathematics, C(n,m) is defined in clever way: it's zero when n,m make no sense (e.g. m<0). Such definition allow simplify much the summation domain. In particular, the summation can be done in Maple as follows:

add(add(add(add(binomial(20,x)*binomial(20,y)*binomial(10,z)*binomial(x,t)*binomial(y,z+t-x)*binomial(z,y-t),t=0..x),z=25-x-y),x=0..20),y=0..20);

producing the result:  19507489009223604480
0

LVL 1

Expert Comment

ID: 10011022
And for 20 gammas, the result is

add(add(add(add(binomial(20,x)*binomial(20,y)*binomial(20,z)*binomial(x,t)*binomial(y,z+t-x)*binomial(z,y-t),t=0..x),z=30-x-y),x=0..20),y=0..20);

1825444985199239330887936

0

LVL 84

Expert Comment

ID: 10016381
Hmm, I got 19507489009223604480 and 1825444985199239090513920.  I must have missed a term for the 20 20 20 case.
The Maple notation is much nicer than my explicit list of terms
0

LVL 50

Author Comment

ID: 10020548
Thanks guys

I was used enjoy engineering mathamatics especially when it was about probabality but this was way out of my experience....

I need to do a bit more digging with this but I'll close this now. Ozo I'll give you the points for the hard work but I'll also post points for relf in a new question, the explanation and the formula was much appreciated

Cheers

Dave
0

Expert Comment

ID: 10065581
To brettdj,
Hi Dave, did you get a result from relf's answer? Can you email me, I had some further thoughts but didn't want to cloud the issue, which may now be irrelevant if relf's anwswer works.
cheers
will49
0

LVL 50

Author Comment

ID: 10067885
Hi will49

I marked it down for future reference but I confess that I haven't used the methods above yet

Once I got two explanations I thought I'd close the question rather than drag it out. You may wish to keep posting in this thread to answer your problem

Regards

Dave
0

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