Proving that k is the number of ways we can order k items

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Homework Help Overview

The discussion revolves around proving that k! represents the number of ways to order k items, with a focus on mathematical induction as a method of proof.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the concept of mathematical induction, questioning how to apply the inductive hypothesis to demonstrate that (k + 1)! is the number of ways to order (k + 1) items. There are attempts to clarify the placement of the (k + 1)th object and how it relates to the existing k objects.

Discussion Status

Several participants have provided insights and suggestions, with some indicating that the reasoning is on the right track. However, there remains a lack of consensus on the clarity and rigor of the arguments presented, particularly regarding the inductive steps.

Contextual Notes

Participants express confusion over the terminology used, particularly regarding the placement of the (k + 1)th object and the application of the inductive hypothesis. There is also mention of the potential for the discussion to become convoluted with multiple perspectives being shared.

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Homework Statement



How do you prove that k! is the number of ways we can order k items?


Homework Equations





The Attempt at a Solution



I tried mathematical induction... but I keep getting stuck on how to show that (k + 1)! is the

number of ways we can order (k +1) items?

Can anyone help? Thanks.
 
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This might not be rigorous enough but it may be enough for you to get started:

Take 1 item. How many ways can you place it into k+1 slots? Now by the induction hypothesis, how many ways can you fill the remaining k slots with the remaining k items?
 
number0 said:
I tried mathematical induction... but I keep getting stuck on how to show that (k + 1)! is the

number of ways we can order (k +1) items?

Can anyone help? Thanks.
So you have k objects that you can "line up" in (k!) different ways. Now add one more object to the collection. What are the different ways in which you can position this new object among the existing line of k objects?
 
Both you guys gave me good help, but I am still stuck.

This is what I got so far...

1) There is k+1 ways to place it into k slots.

How am I suppose to use the inductive hypothesis?
 
number0 said:
Both you guys gave me good help, but I am still stuck.

This is what I got so far...

1) There is k+1 ways to place it into k slots.

How am I suppose to use the inductive hypothesis?
It's not clear what the "it" is that you refer to above, or which step of the inductive process you are at.

I suggest you wipe the slate clean and start at the beginning.

What's the first step in a proof by induction?
 
Gokul43201 said:
It's not clear what the "it" is that you refer to above, or which step of the inductive process you are at.

I suggest you wipe the slate clean and start at the beginning.

What's the first step in a proof by induction?

By it, I am referring to k+1. So there is k+1 ways to place k+1 into a set that contains k elements.
 
By it, I am referring to k+1.
By "k+1", do you mean "the (k+1)th object"?

If so, you are correct, but can you explain more clearly how there are k+1 ways to position this object?
 
Gokul43201 said:
By "k=1", do you mean "the (k+1)th object"?

If so, you are correct, but can you explain more clearly how there are k+1 ways to position this object?

Well... take k = 3 for example. So k + 1 = 4. Then:

1 2 3 4
1 2 4 3
1 4 2 3
4 1 2 3

. Thus there are four ways?
 
Say the (k+1)th object is in the first spot. How many ways are there to arrange objects 1 through k in the remaining spots?
 
  • #10
vela said:
Say the (k+1)th object is in the first spot. How many ways are there to arrange objects 1 through k in the remaining spots?

Oh! k times. Then the logic proof can goes as follows:

since k + 1 can be placed in k amount of positions, we can use the inductive hypothesis to multiply k + 1 by k! to get (k + 1)! ?

I am not sure about it;/
 
  • #11
What you wrote has numerous mistakes, but I suspect it may just be you being sloppy. For example, you said "k + 1 can be placed in k amount of positions," yet in your previous post, you argued that the (k+1)-th item can go into k+1 spots. Can you clarify exactly what you're thinking?
 
  • #12
vela said:
What you wrote has numerous mistakes, but I suspect it may just be you being sloppy. For example, you said "k + 1 can be placed in k amount of positions," yet in your previous post, you argued that the (k+1)-th item can go into k+1 spots. Can you clarify exactly what you're thinking?

Okay I reread my post and it was sloppy. This is what it was suppose to mean:

If we insert a k+1 element into the set, then there will be k + 1 amount of ways to position k + 1. From the inductive hypothesis, we can multiply k + 1 ways by k! to get (k + 1)! ways of arranging a set.

Is this almost right?
 
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  • #13
vela said:
Say the (k+1)th object is in the first spot. How many ways are there to arrange objects 1 through k in the remaining spots?
This is no longer a proof by induction, is it?

I'm going to step out of this thread - don't want this to become a case of too many cooks.

But before I leave, I should perhaps add that the discussion in post #8 is on the right track. All you (number0) need to do is make that argument using general terms, rather than a specific example. And then you are one step away from completing the proof by induction.
 
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  • #14
number0 said:
Okay I reread my post and it was sloppy. This is what it was suppose to mean:

If we insert a k+1 element into the set, then there will be k + 1 amount of ways to position k + 1. From the inductive hypothesis, we can multiply k + 1 ways by k! to get (k + 1)! ways of arranging a set.

Is this almost right?
Yup, that's the idea. You just have to fill in the rest of the pieces of a proof by induction.
Gokul43201 said:
This is not a proof by induction, is it?
Well, it's an element of the proof. The induction hypothesis allows you to assume there are k! ways to arrange the k remaining elements.
I'm going to step out of this thread - don't want this to become a case of too many cooks.
I don't think it's necessary for you to stop posting here. It would be one thing if we were proposing two completely different ways to solve this, but we're essentially saying the same things.
 

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