# Way to find the number of permutations with a specific requirement.

The question is

Let $$(x_1, x_2, ..., x_{15})$$ be a permutation of $$(1, 2, 3, ..., 15)$$ that satisfies $$x_1>x_2>x_3>\dots>x_7\hspace{0.25cm}\text{and}\hspace{0.25cm}x_7 If $$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.

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...
• @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. Commented Nov 17, 2022 at 14:10
• 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. Commented Nov 17, 2022 at 14:14
• Do we multiply by $2$ because of there are $2$ possible numbers for $x_6$ and $x_8$? Commented Nov 17, 2022 at 14:24
• @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. Commented Nov 17, 2022 at 14:27