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bobbym
2013-03-27 15:12:14

Hi;

You are welcome and welcome to the forum.

rhymin
2013-03-27 15:04:13

Thank you!

bobbym
2013-03-27 15:02:35

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

rhymin
2013-03-27 14:57:05

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?

bobbym
2013-03-27 14:49:12

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.

rhymin
2013-03-27 14:42:00

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.

bobbym
2013-03-27 13:52:02

Hi;

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

Can you do it now?

rhymin
2013-03-27 13:49:05

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?