Why is n the number of possible ways to arrange n distinct items?

  • Thread starter Thread starter zeion
  • Start date Start date
Click For Summary

Homework Help Overview

The discussion revolves around understanding why the factorial notation n! represents the number of ways to arrange n distinct items. Participants are exploring the reasoning behind this concept in combinatorics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to understand the multiplication principle in the context of arranging items. Questions are raised about the reasoning behind multiplying the number of choices for each slot as items are placed.

Discussion Status

Some participants have provided examples and suggested practical experimentation with physical objects to illustrate the concept. There is an ongoing exploration of how the multiplication rule applies to the arrangement of items, with no explicit consensus yet.

Contextual Notes

Participants are encouraged to consider small cases and build up to larger numbers of items, indicating a focus on experiential learning. There is mention of the need to clarify definitions related to the multiplication rule.

zeion
Messages
455
Reaction score
1

Homework Statement



Why?

Homework Equations





The Attempt at a Solution



I know that n! = n(n-1)...1
and if n = 3 then the possibility for the first entry is 1 or 2 or 3, and if the first entry is 1 then the second cannot be 1 so the second entry has 2 possibilities. But why do I multiply
 
Physics news on Phys.org
Because n! is literally a generalization from direct human experience. Try this yourself! See how many ways you can fill n slots using n distinct objects; start small, very small, and continue upward. Try with poker chips, marbles, ... anything you want.
 
Think of there being n slots. You want to find all possible arrangements.

How many ways of choosing an item for the first slot? We have n items to choose from, so it must be n.

Once we placed an item in the first slot, how many ways are there of choosing an item (from the remaining items) for the second slot?

After we placed the first two items, how many ways of choosing an item for the third slot?

.

.

.

Once you have them all, you should know the multiplication rule and...
(If you're not sure what the multiplication rule is, look up the definition and think about how it relates to the problem)
 
Say you have 4 objects, A, B, C, and D, and you want to fill two slots. You have these possibilities.

AB AC AD
BA BC BD
CA CB CD
DA DB DC

Each row corresponds to a choice of filling the first slot, and the entries across each row just go through the possibilities of filling the second slot. So you can see the total number of possibilities is just the number of rows times the number of columns. Do you see how it generalizes?
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
3K
Replies
10
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 17 ·
Replies
17
Views
3K
Replies
1
Views
2K
  • · Replies 29 ·
Replies
29
Views
7K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 12 ·
Replies
12
Views
3K