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

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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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,791

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.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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

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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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,791

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.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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

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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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,791

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.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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

Thank you!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,791

Hi;

You are welcome and welcome to the forum.

**In mathematics, you don't understand things. You just get used to them.**

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