Hangout Rearrangements

the first condition "the number of ways that the pictures can be arranged so that my picture is next to Mehrnoush's picture" can be counted as $2(n-1)!$ where n is the total number of students because you will treat Christian's and Mehrnoush's pictures as 1, considering that they must be together. it is multiplied by 2 because you can get the number of arrangements of the 2 pictures which is 2! or simply, 2.

The second condition "there are either 7 or 8 pictures between my picture and Mehrnoush's picture" can be calculated as $2(n-9+1)(n-2)! + 2(n-10+1)(n-2)!$ or for simplification, $(2(n-2)!)(2n-17)$ where n is the number of students. The 2 at the formula is because the 2 pictures can be interchanged. There is $(n-9+1)$ because there will be changes in position of the 2 pictures with 7 pictures in between which will be more as the number of students' pictures increase; same goes with $(n-10+1)$ . $(n-2)!$ is there because the 2 pictures are somewhat fixed there, while the pictures in between and the rest are the only ones which can have different arrangement.

Hence, we have $2(n-1)! = 2 (2n-17)(n-2)!$ , which gives $n-1 = 2n-17$ , or that $n = 16$ .

Sorry for my poor english. Let n the number of students. Two pictures a and b can be arranged in (n-1) ways so that a is next to b. The first one can be placed in position 1,..,n-1. For distance 7 there are only n-8 ways to place the first picture, from 1 to n-8. Similarly, there are n-9 ways to place the first of two pictures so that there are 8 pictures between them. In all cases, for each choise of first position, there are 2x(n-2)! permutations of a,b and other n-2 pictures, so we can consider only the ways to place the first picture in both cases. From n-1=(n-8)+(n-9) we find n=16

Let $n$ be the number of students in the master session. For a given value of $k$ , the number of ways for Christian and Mehrnush's photos to have exactly $k$ photos between them is $2(n-k-1)(n-2)!$ . So the number of ways for them to be consecutive is $2(n-1)(n-2)!$ , the number of ways for them to be 7 or 8 apart is $2(n-8)(n-2)! + 2(n-9)(n-2)! = 2(2n-17)(n-2)!$ . Setting these equal, we get that $2(n-16)(n-2)! = 0$ . Since $n \geq 9$ , $(n-2)! \neq 0$ , so $n = 16$ .