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The question is

Let $(x_1, x_2, ..., x_{15})$ be a permutation of $(1, 2, 3, ..., 15)$ that satisfies $$x_1>x_2>x_3>\dots>x_7\hspace{0.25cm}\text{and}\hspace{0.25cm}x_7<x_8<x_9<\dots x_{15}.$$ If $x_6$ and $x_8$ are either 2 or 3, the number of 15-tuple that satisfy the permutation is...

I noticed that if $x_6=2$ and $x_8=3$, then $x_7$ would have to be $1$ so as to satisfy the requirement. Similarly, if $x_6=3$ and $x_8=2$, then $x_7$ would still be $1$. Then, counting the number of possible $x_5$'s, bearing in mind the $x_1, x_2, x_3, x_4$, there are 8 choices for $x_5$ which are $4$ to $11$.

I'm stuck here and I don't know how to advance. Any help would be much appreciated, thanks.

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It should be clear that $x_7$ must equal $1$ since it is smaller than everything else. Next, pick which of $x_6$ or $x_8$ is equal to $2$ and let the other be equal to $3$. Now... consider... If we were to take five other unused numbers, could we uniquely and unambiguously find a specific order in which we place those as $x_1$ through $x_5$ and the remaining unused numbers as $x_9$ through $x_{15}$? Yes? And any valid permutation of your type we could have uniquely and unambiguously talked about the set consisting of the first five numbers and then a selection of either $2$ or $3$? Yes? Good. It follows then that the number of possible permutations is going to be ____.

$2\cdot \binom{12}{5}$

For examples: If the set of five chosen was $\{4,5,6,7,8\}$ and we chose the number $2$ then the valid permutation would have been $(8,7,6,5,4,2,1,3,9,10,11,12,13,14,15)$. On the other hand if the set of five chosen were $\{8,10,12,13,14\}$ followed by $3$ the corresponding permutation would have been $(14,13,12,10,8,3,1,2,4,5,6,7,9,11,15)$ and so on...

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  • $\begingroup$ @N.F.Taussig If that was actually supposed to be a part of the condition of the problem, then the approach is easily modified. Just decide which is $2$ and which is $3$. Then approach the same way, picking what subset of size five this time goes for the first five positions in decreasing order. $\endgroup$
    – JMoravitz
    Commented Nov 17, 2022 at 14:10
  • $\begingroup$ Ah okay, thanks! It is actually a requirement of the problem: if $x_6$ is $2$, then $x_8$ must be $3$, and vice versa. Sorry for the bad wording that eventually caused this misunderstanding. $\endgroup$
    – ryan.zcd
    Commented Nov 17, 2022 at 14:14
  • $\begingroup$ Do we multiply by $2$ because of there are $2$ possible numbers for $x_6$ and $x_8$? $\endgroup$
    – ryan.zcd
    Commented Nov 17, 2022 at 14:24
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    $\begingroup$ @nich because there are two possible scenarios: either $x_6=2,x_8=3$ or that $x_6=3,x_8=2$. If you wanted to phrase it that way, sure. I would emphasize that the choice for one forces the choice for the other however. $\endgroup$
    – JMoravitz
    Commented Nov 17, 2022 at 14:27

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