Possible number of combinations.

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Discussion Overview

The discussion revolves around the number of possible combinations when rearranging a set of objects, particularly when certain objects are restricted from specific positions. Participants explore the implications of these restrictions on the total number of arrangements, considering both single and multiple constraints.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant states that the total number of arrangements for n objects is n!, and questions how the removal of one object from the first position affects the total combinations.
  • Another participant agrees that removing one object from the first position results in (n-1)*(n-1)*(n-2)*(n-3)... but does not clarify how this relates to other positions.
  • There is a proposal that if an object cannot be placed in any position, the total combinations remain (n-1)*(n-1)*(n-2)*(n-3)...
  • A participant suggests that if there are n types of objects available in infinite quantity, and one type cannot occupy a specific place, the combinations would be (n-1)*n*n*n... and questions the outcome with multiple restrictions.
  • Another participant confirms the mathematical manipulation presented regarding the removal of combinations due to constraints, but does not provide a definitive resolution to the implications of these constraints.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and agreement on the implications of constraints on combinations, with some confirming calculations while others question the reasoning behind certain assumptions. The discussion does not reach a consensus on the overall impact of these restrictions.

Contextual Notes

Some assumptions regarding the nature of the objects and the specific constraints are not fully articulated, leading to potential ambiguity in the reasoning presented.

dE_logics
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I just wanted to confirm this...suppose we need to rearrange n number of objects, the possible ways they can be rearranged is n!

If suppose at the first place, one of the objects cannot be admitted...then it will be (n-1)*(n-1)*(n-2)*(n-3)...right?

Instead of the first place, suppose, an object cannot be admitted to any other place...it won't make any difference right?...I mean, the total combinations will still be (n-1)*(n-1)*(n-2)*(n-3)?

Also related to the above question, if we have n 'types' of objects infinite in quantity (actually it's more of permutation) and one of them cannot be admitted to a place (not necessarily the first)...then the possible number of combinations will be (n-1)*n*n*n...etc...right? Also in case of multiple exceptions for instance object 'x' cannot be admitted to first, second and third place or object 'x' cannot be admitted to first, object 'y' cannot be admitted to second and finally object 'z' cannot be admitted to the third place, then will the possible combinations will be (n-1)*(n-1)*(n-1)*n*n*n... etc ?
 
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Is it that no one is understanding the question?
 
dE_logics said:
I just wanted to confirm this...suppose we need to rearrange n number of objects, the possible ways they can be rearranged is n!

If suppose at the first place, one of the objects cannot be admitted...then it will be (n-1)*(n-1)*(n-2)*(n-3)...right?
Yes. The total number of combinations without the constraint is n!. The constraint removes (n-1)! combinations so the total number is:
n! - (n-1)! = n*(n-1)! - (n-1)! = (n-1)(n-1)! = (n-1)(n-1)(n-2)...*2*1
Instead of the first place, suppose, an object cannot be admitted to any other place...it won't make any difference right?...I mean, the total combinations will still be (n-1)*(n-1)*(n-2)*(n-3)?
How could it matter?

Also related to the above question, if we have n 'types' of objects infinite in quantity (actually it's more of permutation) and one of them cannot be admitted to a place (not necessarily the first)...then the possible number of combinations will be (n-1)*n*n*n...etc...right? Also in case of multiple exceptions for instance object 'x' cannot be admitted to first, second and third place or object 'x' cannot be admitted to first, object 'y' cannot be admitted to second and finally object 'z' cannot be admitted to the third place, then will the possible combinations will be (n-1)*(n-1)*(n-1)*n*n*n... etc ?
correct
 
Last edited:
n! - (n-1)! = n*(n-1)! - (n-1)! = (n-1)(n-1)! = (n-1)(n-1)(n-2)...*2*1

So I'm right?

How could it matter?

Ok, thanks.

correct

Again...thanks.
 

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