Permutations of (abc)(efg)(h) in S7

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The discussion focuses on calculating the distinct permutations of the form (abc)(efg)(h) in S7. Participants analyze the selection process, initially suggesting to use combinations and then addressing the issue of overcounting due to equivalent arrangements. It is clarified that the permutations (abc)(efg)(h) and (efg)(abc)(h) are identical, leading to the need for division by 2! instead of 3!. The reasoning centers on the fact that once the sets abc and efg are chosen, the last element h does not contribute to additional arrangements. The conversation concludes with an understanding that the method of selection avoids double counting of the arrangements.
cragar
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Homework Statement


How many distinct permutations are there of the form (abc)(efg)(h) in S7?

Homework Equations


3. The Attempt at a Solution [/B]
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.
 
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cragar said:

Homework Statement


How many distinct permutations are there of the form (abc)(efg)(h) in S7?

Homework Equations


3. The Attempt at a Solution [/B]
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.
Is the permutation (abc)(efg)(h) different from the permutation (efg)(abc)(h)?
 
no, so I then need to divide by 3!
 
cragar said:
no, so I then need to divide by 3!
How do you arrive at 3?
 
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.
 
cragar said:
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.
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?
 
oh your saying only divide by 2! because by the time we choose our last element h we only have one choice, so it won't affect the counting.
Because by the time (abc)(efg) have been chosen we only have one choice left.
 
cragar said:
oh your saying only divide by 2! because by the time we choose our last element h we only have one choice, so it won't affect the counting.
Because by the time (abc)(efg) have been chosen we only have one choice left.
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.
 
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