Different Combinations of text

[pauldaddyadams]

Hi,

I have a list of 5 words and I want to know the different combinations of these words.

The List is:
Stp (Yes/No)
Shp (Yes/No)
Ld (Yes/No)
Geo (Yes/No)
Gas (Yes/No)


e.g. for Stp
Stp Shp Ld Geo Gas
Yes No No No No
Yes Yes No No No
Yes No Yes No No
Yes No No Yes No
Yes No No No Yes
Yes Yes Yes No No
Yes Yes No Yes No
Yes Yes No No Yes

And so on......

I have found the attached spreadsheet that may do it but despite trying I couldnt get it to list all combinations.

Can anyone get the attached to help solve my problem or point me in the direction of a simlar model? (Original post: http://www.excelforum.com/excel-misc...ent-words.html)

Paul

[MarvinP]

Hi Paul,

The words Combination and Permutation are often confused to us laymen. First you need to know if the order of the words is a different and both count or not. The next question would be to calculate how many rows the answer will have. Read:
http://www.mathsisfun.com/combinator...mutations.html
to answer the above.

[pauldaddyadams]

Hi,

It would be a combination - the order doesnt matter.

Repetition is not allowed.

i still cant work out how many different combinations I am expecting :-/

[MarvinP]

Sooooooo.

If the order doesn't matter then
Yes, Yes, Yes, Yes, No is the same as Yes, No, Yes, Yes, Yes! Is that correct?

Now what does Stp, Shp, Ld, Geo and Gas have to do with this problem?

I still don't understand the question.

Here is why! I used to teach high school math. In Advanced Algebra we had a section on Combinations and Permutations. The killer problem was to put 4 black balls and 3 white balls in a bag. You would reach in and pull out a ball and see what color it was. In one type of problem you would put the ball back in the bag and draw again (with replacements). In another type of problem you would keep the ball and draw from what was left (no replacements). At the end of chapter test they would always ask, "Take 4 balls from the bag" and not mention replacements or not. I always had to look at the answer in the back of the book to see if they put those suckers back before drawing a second time or if they drew them all at once. Sometimes they meant one at a time with replacements which is much different than all at a time (no replacements).

Defining the question to understand how the game is played is the most important part of these kinds of problems.

[pauldaddyadams]

MarivnP – hope your patient! Lol

This has had me and my colleague in a twist – its been a long time since I have had to think about the number of combinations.

Previously I told you it doesn’t matter however this must be incorrect based on your example, so order does matter.

Stp, Shp, Ld, Geo and Gas can either be Yes or No. In my new example I have called them:
Stp and Unstp
Shp and Unshp
Ld and UnLd
Geo and UnGeo
Gas and UnGas


So for example you can have the following
Stp, Shp, Ld, Geo, Gas
Stp, UnShp, Ld, Geo, Gas etc

In the attached I think it lists all combinations however I am not confident at all. Can you confirm it?

[MarvinP]

This is starting to make more sense to me.

Lets say you have 5 cards in a deck. You deal the first card and you have 4 left to deal (no replacements)
The number of hands is
5 factorial or 5 possible first cards (times) 4 possible second cards (times) 3 times two time 1. (order matters)

But what about the Stp vs UnStp???? How about you can put the card face up or face down. That might simulate your problem.

So 5 for the first card time 2 as it might go face up or down.
Then 4 for second card times 2

How about
5*2*4*2*3*2*2*2*1*2 for all possible ways to do this problem.
Total = 5 factorial times (2^5) = 120 * 32 = 3840 different ways to deal that deck using your rules.

Is this the correct assumptions on what you want?

[pauldaddyadams]

I found this on a different forum which helps explain it better than I can:

Original Question - (similar to mine 5 combinations, yes or no)
"Say I have 5 patents going into trial. All, one, none, or any combination of them may be found to infringe (so 2 options per patent - either it infringes or it doesn't)

How many possible outcomes/combinations are there?"

Answer
This question deals with combinations (not permutations), since order doesn't matter here (Patent #1 and Patent #2 both infringing is the same as Patent #2 and Patent #1 infringing).

This is actually the sum of 6 different combinations.
1) the combination where none of 5 patents infringes (n=5, k=0)
2) 1 of 5 infringes (n=5, k=1)
3) 2 of 5 (n=5, k=2)
4) 3 of 5 (n=5, k=3)
5) 4 of 5 (n=5, k=4)
6) 5 of 5 (n=5, k=5)

So the total number of combinations is

C(5,0)+C(5,1)+C(5,2)+C(5,3)+C(5,4)+C(5,5)

=1+5+10+10+5+1=32


Do you agree with this? I believe 32 is correct which is what the attachement shows

[pauldaddyadams]

Anyone know the answer? I've been searching online and I'm still not sure

[pauldaddyadams]

I found this site: (Universal Text Combination Generator section)
http://textmechanic.com/Permutation-Generator.html

It only goes up to 4 different combinations which total 16 different combinations but based on this I am confident 5 will have 32