# Question on HW 3, Q1(b)

If I’m understanding the definition of a first run correctly, then the lengths of the first runs of the permutations of $[3]$ are as follows:

123:  3

132:  2

213:  2

231:  2

312:  1

321:  1

Thus, the average length of a first run is $11/6$, but $1/1! + 1/2! + 1/3! = 5/3 \ne 11/6$. Am I doing something wrong?

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### 8 comments

1. I think the permutation 213 has 1 as the length of its leftmost run. The 2 by itself, yes? This should give $10/6$ for the average length, which does match up.

• dgklein

Isn’t 23 the leftmost run of 213? It doesn’t seem like the $i_1, \ldots, i_k$ in the definition are required to be consecutive.

2. suitored

@dgklein But then how do you distinguish “leftmost” run from any other run? I think a leftmost run must begin at the beginning, right?

• suitored

@dgklein Sorry, I think I misunderstood your point…now I’m confused too

3. suitored

I did it for $n = 4$ and it seems they have to be consecutive…

4. Sorry guys, I wrote HW3 on an airplane, and apparently I wasn’t writing very lucidly. I already fixed the mistake on the website.
In problem 1(b), a run in a permutation $w$ is a maximal sequence of increasing **consecutive** elements $w_i < ... < w_{i+k}$. So the runs of 213 are 2 and 13.

On the other hand, I stated problem 1(d) slightly vaguely (what if there is no $m$th run? Etc.). I did that on purpose. It's good for you to get used to slightly open-ended question. 🙂

5. So, for the example of 213, the leftmost run is 2?

• Anastasia, that’s correct.