Permutations of a group

1. Feb 8, 2016

cragar

1. The problem statement, all variables and given/known data
How many distinct permutations are there of the form (abc)(efg)(h) in S7?
2. Relevant equations
3. The attempt at a solution

since we have 7 elements I think for the first part it should be 7 choose 3 then 4 choose 3.
And then we multiply those together.

2. Feb 8, 2016

Samy_A

Is the permutation (abc)(efg)(h) different from the permutation (efg)(abc)(h)?

3. Feb 8, 2016

cragar

no, so I then need to divide by 3!

4. Feb 8, 2016

haruspex

How do you arrive at 3?

5. Feb 8, 2016

cragar

i mean 3 factorial, because (abc)(efg)(h)= (efg)(abc)(h) so I need to divide by 6 because their are 6 ways to arrange that and we would be over counting.

6. Feb 8, 2016

haruspex

Yes, I understood it was factorial, but I was specifically challenging the 3.
I only see two equivalent arrangements there. What other four have you redundantly counted by your method?

7. Feb 9, 2016

cragar

oh your saying only divide by 2! because by the time we choose our last element h we only have one choice, so it wont affect the counting.
Because by the time (abc)(efg) have been chosen we only have one choice left.

8. Feb 9, 2016

haruspex

Yes, 2, but not for that reason.
You chose abc as a set of three, so you covered all permutations of those three in one selection - no double counting so far.
Likewise in choosing the set def.
The double counting arises because you could have chosen the set def first, then the set abc.