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## #1 2013-03-26 14:49:05

rhymin
Member
Registered: 2013-03-26
Posts: 20

### Permutation

A bit confused on how to begin this.

Consider the permutation of 1, 2, 3, 4. The permutation 1432, for instance, is said to have one ascent  namely, 14 (since 1 < 4). This same permutation also has two descents  namely, 43 (since 4 > 3) and 32 (since 3 > 2). The permutation 1423, on the other hand, has two ascents, at 14 and 23  and the one descent 42.

a) How many permutations of 1, 2, 3 have k ascents, for k = 0, 1, 2?

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## #2 2013-03-26 14:52:02

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

### Re: Permutation

Hi;

Did you try writing down all the permutations, there are only 6?

Can you do it now?

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #3 2013-03-26 15:42:00

rhymin
Member
Registered: 2013-03-26
Posts: 20

### Re: Permutation

I guess what confused me the most was the last part of the question, "for k = 0, 1, 2".  What exactly does this mean?  Because the example right before it doesn't mention anything about that.

Thank you for your quick reply!

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## #4 2013-03-26 15:49:12

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

### Re: Permutation

Look at the first one 1,2,3

2 >1 so that is an ascent, then 3 > 2 that is another ascent. So 1,2,3 has 2 ascents.

Now look at 3,2,1. 3 is not less than 2 and 2 is not less than 1. 3,2,1 has no ascents.

Then just want you to look at all 6 and find the ones with 0 ascents, 1 ascent and 2 ascents.

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #5 2013-03-26 15:57:05

rhymin
Member
Registered: 2013-03-26
Posts: 20

### Re: Permutation

Ohh, thank you for that, I get it now.
123: 2 ascents
132: 1 ascent
213: 1 ascent
231: 1 ascent
312: 1 ascent
321: 0 ascents

How would you write the answer? Just like this?

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## #6 2013-03-26 16:02:35

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

### Re: Permutation

Say, the premutations of (1,2,3) have 1 zero ascent, 4 ascents of 1 and 1 ascent of 2.

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #7 2013-03-26 16:04:13

rhymin
Member
Registered: 2013-03-26
Posts: 20

Thank you!

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## #8 2013-03-26 16:12:14

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

### Re: Permutation

Hi;

You are welcome and welcome to the forum.

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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