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Suppose you have a binary sequence $s_t$ of length T. We transform this sequence in integers by replacing the zeroes with 1,2,3….,k and the ones by k+1,k+2….T.

For example 000000 is transformed in 123456 and 100000 is transformed in 612345 and finaly 101010 is 415263.

Now i would like to search for patterns in the transformed sequence, i grab subsequences of lenght 3 (m=3) and:

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- if $x1<x2<x3$ then this is pattern 123
- if $x1<x3<x2$ then this is pattern 132
- if $x2<x1<x3$ then this is pattern 213
- if $x2<x3<x1$ then this is pattern 231
- if $x3<x1<x2$ then this is pattern 312
- if $x3<x2<x1$ then this is pattern 321

*for example 101010 = 415236
415 = 213;
152 = 132;
523 = 231;
236 = 123*

But opcion 6 (321) is never going to happen for the nature of the trasformation.

**How many patterns are there?** In this case the answer is 5 (all the listed but option six could happen)

If you want a subsequence for a greater m the answer is $2^m-m$ I do not understand how to arrive to this answer. If anyone could help me!

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The admissible patterns are exactly the image of your transformation for a sequence of length of length $m$. i.e two interleaved increasing sequences.

So to count them you merely need to count the number of ways to insert the sequence $1, \ldots, k$, which is $\binom{m}{k}$ ($2^m$ in total (including the empty sequence (k=0))). However, if $1, \ldots, k$ is at the beginning then it gives the same sequence so you subtract the $m$ duplicates, giving $2^m -m$.

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