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I have a question that I need for a larger project I'm working on. I don't know how to precisely explain it (hence I could not search it on the internet) but I'll do my best. First consider the list of indices $\{1,2,3,...,n\}$, and permutations of these. There are n! unique permutations.

Now suppose we divide this into two groups, where the first group ends somewhere before n. In other words, $\{1,2,3,...,n\} = \{1,2,3,...,n_1,n_1+1,n_1+2,...n\}$, with group 1 being $\{1,2,...,n_1\}$ and group 2 being $\{n_1+1,...,n\}$. My question is if we separate the groups while respecting the within group order, how many occurrences are there of a certain permutation in each group?

For example, take permutations of $\{1,2,3,4,5,6\}$. Take group 1 to be $\{1,2,3,4\}$ and group 2 to be $\{5,6\}$. Now consider a certain permutation of the groups, e.g. $\{2,3,4,1\}\{6,5\}$. Then for example, the following permutations would be what I am looking for: $\{2,3,4,1,6,5\}$ or $\{2,6,5,3,4,1\}$ or $\{6,2,3,4,1,5\}$ etc. Each of these works because if I were to 'erase' the group 2 terms I would get $\{2,3,4,1\}$ and if I were to erase the group 1 terms I would get $\{6,5\}$. My question is how many of these are there in general? In the case of this example there are at least 3.

Thank You!

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Pick which positions are occupied by the first group. Now that you have picked the positions, simply insert the specific permutation into those positions. In the remaining unused positions, insert the specific permutation of the second group. There are then $\binom{n}{n_1}$ permutations satisfying the conditions you describe.

In your example, there are $\binom{6}{4}=15$ such permutations where the elements $\{2,3,4,1\}$ appear in that order and the elements $\{6,5\}$ appear in that order (possibly interwoven).

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    Couldn't you make a similar argument for \binom{n}{n_2} by choosing the positions for group 2 first? – 2017-02-21
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    Yes, if you were to have actually written the number $n_2$ anywhere to begin with. Since you actually implied the number $n_1$ was known and made no mention of the any number called $n_2$ I opted use use $n_1$ in my argument instead of creating an additional unnecessary variable that didn't before exist. Of course $\binom{n}{r}=\binom{n}{n-r}$ is a well known combinatorial identity. – 2017-02-21