1. Combinations question

I have a strange question I hope someone can explain it to me or tell me if there is some type of statistical definition for this process im trying to do, ill try to explain it in steps

1. there are 20 possible items to choose from
2. you have to choose 5 of these items, the order does not matter
3. these items all have pairs, so a goes with b, c goes with d. This means if you were to choose A then B is no longer an option.

Now with this info, how many different possible combinations are there? i know with 20 possible choices and having to choose 5, there are 15,504 possible combinations, but there is much less with the other info i gave. How would i figure this out? and is there some type of statistical way of describing this type of combination problem? thank you for any info and sorry if it is confusing or makes no sense

2. Re: Combinations question

If I understand correctly, there are in effect ten separate pairs to choose from, and from each pair one partner can be chosen. That is, there are 20 possibilities for your first choice, 18 for your second, 16 for your third, etc.

From there, you should be able to work out the answer, not forgetting that there are several different ways (how many?) to end up with the same collection of five items.

3. Re: Combinations question

Originally Posted by Con-Tester
If I understand correctly, there are in effect ten separate pairs to choose from, and from each pair one partner can be chosen. That is, there are 20 possibilities for your first choice, 18 for your second, 16 for your third, etc.

From there, you should be able to work out the answer, not forgetting that there are several different ways (how many?) to end up with the same collection of five items.
that is correct, and I have no idea how to work this out which is why I came here to find out if some light can be shed, I am not a statistics person in any way, just trying to find some direction in how to figure this out

4. Re: Combinations question

You have combinations without replacement and order does not matter, conditional on the pairing. So as CT mentioned, with every draw you lose two from the sample. Check this out for some background:

http://www.mathsisfun.com/combinator...mutations.html

5. Re: Combinations question

Okay, so let’s take this one step at a time. Try to answer the following two questions:

1) If you have 20 possibilities for your first choice, 18 for your second, 16 for your third, 14 for your fourth and 12 for your final one, what is the total number of unique sequences that are possible?

2) In how many unique orders can five different items be arranged?

For each question, think of five boxes in a row into each of which you place one item.

6. Re: Combinations question

If A and B are paired would you consider the following two sequences equivalent or would you want to count them seperately:

A C E G H

vs

B C E G H

Are those two different sequences or are they equivalent since A and B are paired?

7. Re: Combinations question

Originally Posted by Dason
If A and B are paired would you consider the following two sequences equivalent or would you want to count them seperately:

A C E G H

vs

B C E G H

Are those two different sequences or are they equivalent since A and B are paired?
pairs would be considered separate so these are 2 different combinations

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