Combinations/Permutations help?

[playingwithnumbrs]

I have 20 values (A-T) that can be arranged into 3 slots in different combinations. Order only matters when the value in the middle slot is different. For example: A-B-C is the same as C-B-A and would count as one combination, but they are both separate from B-A-C, which would count as a second combination because the middle value is different. I need to find how many combinations can be made. What equations would I use?

 

[pka]

I have 20 values (A-T) that can be arranged into 3 slots in different combinations. Order only matters when the value in the middle slot is different. For example: A-B-C is the same as C-B-A and would count as one combination, but they are both separate from B-A-C, which would count as a second combination because the middle value is different. I need to find how many combinations can be made.

Here is your set \(\displaystyle \{A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T\}\) There are \(\displaystyle ^{20}\mathcal{P}_3=20*19*18=6840\) three-element strings from that set. However, there are \(\displaystyle 18\) of those you have excluded. I assume the excluded strings are consecutive.
Therefore \(\displaystyle A-K-S~\&~S-K-A\) are different because they are not consecutive. Please confirm or correct.

 

[LCKurtz]

So ABC and CBA are the same. I assume that ABC and ABD are different? If I understand correctly, it looks like you have 20 ways to pick the middle letter and [MATH]\binom{19}{2}[/MATH] ways of picking the pair of adjacent letters to go with it (assuming they must be different. Must they?).

 

[playingwithnumbrs]

Here is your set \(\displaystyle \{A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T\}\) There are \(\displaystyle ^{20}\mathcal{P}_3=20*19*18=6840\) three-element strings from that set. However, there are \(\displaystyle 18\) of those you have excluded. I assume the excluded strings are consecutive.
Therefore \(\displaystyle A-K-S~\&~S-K-A\) are different because they are not consecutive. Please confirm or correct.

No, they don't need to be consecutive, sorry. A-K-S and S-K-A would be the same.

 

[playingwithnumbrs]

So ABC and CBA are the same. I assume that ABC and ABD are different? If I understand correctly, it looks like you have 20 ways to pick the middle letter and [MATH]\binom{19}{2}[/MATH] ways of picking the pair of adjacent letters to go with it (assuming they must be different. Must they?).

Yes, they must be different. How would that translate into an equation?

 

[pka]

No, they don't need to be consecutive, sorry. A-K-S and S-K-A would be the same.

Thank you for the clarification. Be sure that you always make conditions clear.
Look at reply #3, There are \(\displaystyle 20\) ways to have a middle term.
There are \(\displaystyle \binom{19}{2}=171\) two element subsets of the remaining nineteen letters.
So there are \(\displaystyle 20*171=3420\) possible triples.

 

[playingwithnumbrs]

Thank you for the clarification. Be sure that you always make conditions clear.
Look at reply #3, There are \(\displaystyle 20\) ways to have a middle term.
There are \(\displaystyle \binom{19}{2}=171\) two element subsets of the remaining nineteen letters.
So there are \(\displaystyle 20*171=3420\) possible triples.

Okay, thank you for your help!

 

[Harry_the_cat]

I think the easiest way is to calculate how many 3 different letter permutaions there are, ie 20x19x18 or 20P3 = 6840. But every one of these has a "partner" ie ABC and CBA are to be counted as 1. So divide by 2 to get 3420.