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

In summary, the number of distinct permutations of the form (abc)(efg)(h) in S7 is 7 choose 3 multiplied by 4 choose 3, divided by 2. This is because we have 7 elements and we first choose a set of 3, then a set of 3, and finally a single element, but we need to divide by 2 to account for the double counting of ordering the sets.
  • #1
cragar
2,552
3

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.
 
Physics news on Phys.org
  • #2
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)?
 
  • #3
no, so I then need to divide by 3!
 
  • #4
cragar said:
no, so I then need to divide by 3!
How do you arrive at 3?
 
  • #5
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
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?
 
  • #7
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.
 
  • #8
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.
 
  • Like
Likes Samy_A

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

1. What is a permutation of a group?

A permutation of a group is a rearrangement of the elements in a group. It is a way of changing the order of the elements without changing the group itself.

2. How many permutations are there in a group?

The number of permutations in a group is equal to the factorial of the number of elements in the group. For example, if a group has 5 elements, there are 5! = 5 x 4 x 3 x 2 x 1 = 120 permutations.

3. How do you calculate the number of permutations in a group with repeating elements?

If a group has n elements, with k1 elements of one type, k2 elements of another type, and so on, the number of permutations is given by n! / (k1! * k2! * ... * kn!). This formula takes into account that not all permutations will be unique due to the repeating elements.

4. What is the difference between a permutation and a combination?

A permutation involves rearranging the elements in a group, while a combination does not consider the order of the elements. For example, the permutations of the letters "ABC" would include "ABC", "ACB", "BAC", etc. while the combinations would only include "ABC", "ACB", and "BAC".

5. How are permutations used in real life?

Permutations have many practical applications, such as in coding and cryptography, where they are used to generate unique combinations of characters. They are also used in statistics to calculate the number of possible outcomes in a sample space. Additionally, permutations are used in various fields of science, such as genetics, to analyze the different ways in which genes can be arranged.

Similar threads

  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
3K
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
3K
  • Calculus and Beyond Homework Help
Replies
1
Views
942
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Linear and Abstract Algebra
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
979
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
Back
Top