Thread: Math Question

08012005, 04:05 PM #1Del Rio Guest
Math Question
Ok I know someone on this site must be a math genious. So here goes. I am in charge of making a work list that will run for six months.
There are 12 guys, who will be working in groups of 2. And they will need to be in a different group every month (not grouped with the same people twice.)
That is fine and dandy, but the hospital has 6 floors, so not only do I need every group to be different each month I need each person to get the opportunity to work on each floor.

I was thinking I would let each letter of the alphabet equal a person so you have
A B C D E F G H I J K LREPRESENTS EACH PERSON 12 TOTAL.
So if you draw a grid out 6months down the side the floors over the top. and you start grouping the letters together.....AB, CD......so on it equals a huge pain in the azz. I have gotten 5 months done and 5 floors where no one is grouped together twice, and no one is on the same floor. But it is looking like there is only one possible solution to the problem.
So does anyone know an equation, a web site for an equation, or anything. $50.00 reward for the correct answer or a source that can help me get the right answer.

08012005, 04:18 PM #2
Re: Math Question
..... wow thats pretty tough.
A+B+C=Nothing I know how to do, math is not for me.
I actually tried this, I made a microsoft excel spread sheet, and tried doing it like that, but no luck.

08012005, 04:20 PM #3
Re: Math Question
"Del Rio" wrote:
Ok I know someone on this site must be a math genious. So here goes. I am in charge of making a work list that will run for six months.
There are 12 guys, who will be working in groups of 2. And they will need to be in a different group every month (not grouped with the same people twice.)
That is fine and dandy, but the hospital has 6 floors, so not only do I need every group to be different each month I need each person to get the opportunity to work on each floor.

I was thinking I would let each letter of the alphabet equal a person so you have
A B C D E F G H I J K LREPRESENTS EACH PERSON 12 TOTAL.
So if you draw a grid out 6months down the side the floors over the top. and you start grouping the letters together.....AB, CD......so on it equals a huge pain in the azz. I have gotten 5 months done and 5 floors where no one is grouped together twice, and no one is on the same floor. But it is looking like there is only one possible solution to the problem.
So does anyone know an equation, a web site for an equation, or anything. $50.00 reward for the correct answer or a source that can help me get the right answer.
OK, this is right up my alley, being a mathematics major. I will work on it for you when I have a chance.

08012005, 04:39 PM #4Del Rio Guest
Re: Math Question
"Mr Anderson" wrote:
..... wow thats pretty tough.
A+B+C=Nothing I know how to do, math is not for me.
I actually tried this, I made a microsoft excel spread sheet, and tried doing it like that, but no luck.
Muchluv, I appreciate it man.

08012005, 04:45 PM #5Starter
 Join Date
 Dec 1969
 Posts
 204
Re: Math Question
i gave it a shot, no luck.

08012005, 05:01 PM #6ProBowler
 Join Date
 Dec 1969
 Posts
 444
Re: Math Question
What you have here is an exercise in logic (propositional or predicate logic, i forgot the difference) so excel is going to be a huge waste of time. You can write all the IF and NOT relationships into the formula, but it's going to be trial and error (probably more error than you want) and cause for a huge headache.
If you want to use an MS office computer program, use Access. Its scripting language, SQL, is appropriate to this relational problem. You can use SQL place logical conditions on the relationships between different cases (i.e., A, B, C...). I'm certain it can be done, but the time investment in it might be more than doing it by hand (but probably less than using excell's formula thing). But if you have to do this sort of exercise again, heck it probably is worth coming up with a logical expression that will sort out all the relationships you need to create. At the very least, using Access just to create your matrix might be a huge help too.
If you're not a fan of learning Access, you might want to search SQL in google and see if there's a site where all the SQL brainiacs share their solutions to questions like yours.

08012005, 05:06 PM #7ProBowler
 Join Date
 Dec 1969
 Posts
 391
Re: Math Question
There are 66 different groups of two that you can form from 12 people. Its called Pascal's Triangle, and the formula is =(n!/((nr)!*r!)) where n is the number of articles, and r is the combinations. In your case, it would go as follows:
(12!/((122)!*2!))
(479001600/((362880)*2))
479001600/7257600=66
Your triangle of combinations looks like this:
AB BC CD DE EF FG GH HI IJ JK KL
AC BD CE DF EG FH GI HJ IK JL
AD BE CF DG EH FI GJ HK IL
AE BF CG DH EI FJ GK HL
AF BG CH DI EJ FK GL
AG BH CI DJ EK FL
AH BI CJ DK EL
AI BJ CK DL
AJ BK CL
AK BL
AL
You have 66 combinations to place in your 36 slot spreadsheet (6 months * 6 floors) with no one working with the same person twice. With a little manipulation, you should be able to get everyone to work every floor twice. If I have some free time at work, I will work on it some more. Untill then, you can google "Pascal's Triangle" for more info...
Hope this helps.

08012005, 05:19 PM #8
Re: Math Question
Tough one.
Caine

Re: Math Question
"Del Rio" wrote:
Ok I know someone on this site must be a math genious. So here goes. I am in charge of making a work list that will run for six months.
There are 12 guys, who will be working in groups of 2. And they will need to be in a different group every month (not grouped with the same people twice.)
That is fine and dandy, but the hospital has 6 floors, so not only do I need every group to be different each month I need each person to get the opportunity to work on each floor.

I was thinking I would let each letter of the alphabet equal a person so you have
A B C D E F G H I J K LREPRESENTS EACH PERSON 12 TOTAL.
So if you draw a grid out 6months down the side the floors over the top. and you start grouping the letters together.....AB, CD......so on it equals a huge pain in the azz. I have gotten 5 months done and 5 floors where no one is grouped together twice, and no one is on the same floor. But it is looking like there is only one possible solution to the problem.
So does anyone know an equation, a web site for an equation, or anything. $50.00 reward for the correct answer or a source that can help me get the right answer.

08012005, 05:26 PM #10Del Rio Guest
Re: Math Question
"ryanmurphy" wrote:
There are 66 different groups of two that you can form from 12 people. Its called Pascal's Triangle, and the formula is =(n!/((nr)!*r!)) where n is the number of articles, and r is the combinations. In your case, it would go as follows:
(12!/((122)!*2!))
(479001600/((362880)*2))
479001600/7257600=66
Your triangle of combinations looks like this:
AB BC CD DE EF FG GH HI IJ JK KL
AC BD CE DF EG FH GI HJ IK JL
AD BE CF DG EH FI GJ HK IL
AE BF CG DH EI FJ GK HL
AF BG CH DI EJ FK GL
AG BH CI DJ EK FL
AH BI CJ DK EL
AI BJ CK DL
AJ BK CL
AK BL
AL
You have 66 combinations to place in your 36 slot spreadsheet (6 months * 6 floors) with no one working with the same person twice. With a little manipulation, you should be able to get everyone to work every floor twice. If I have some free time at work, I will work on it some more. Untill then, you can google "Pascal's Triangle" for more info...
Hope this helps.
Is there only one possible solution? Or are there multiple solutions?
The Closest I have come is
AB CD EL GH FI JK
CE FK HJ BI DG AL
DF AE GK CJ BL (*)
GJ BH AI FL CK DE
HI GL BD EK AJ (+)
KL IJ CF AD EH BG
(*)On month 3 floor 6 the only possible solution is HI which has been used.
(+)On Month 5 floor 6 the only possible solution is CF which has been used.
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