Number of permutations of {1, 2, . . . , n}, in which 1 and 2 are next to each other, with 1 to the left of 2 Let n $\geq$ 4 be an integer. Determine the number of permutations of
$\{1, 2, . . . , n\}$, in which $1$ and $2$ are next to each other, with $1$ to the left of $2$.
I can't make sense of this problem statement. The way I see it, if $n$ is an integer, then the pair $1,2$ could be formed by any pair with the form $\overline{...x_{i-2}x_{i-1}x_i1}, \overline{2y_{1}y_2y_3...}$ or a number with the form $\overline{...x_{i-2}x_{i-1}x_i12x_{i+1}x_{i+2}x_{i+3}..}$ with $x$'s and $y$'s are some mysterious digits. Can anyone explain this problem?
 A: Think of "12" as a single object, say "a".  Then the problem becomes to determine the number of permutations of the n- 1 objects, "a, 3, 4, ..., n".  There are (n-1)! such permutations.
A: The problem statement requires that we count the number of arrangements such that


*

*The arrangement is a permutation of $\{1,2,\dots,n\}$ (i.e. it uses each and every number from $\{1,2,\dots,n\}$ exactly once)

*$1$ and $2$ are adjacent, i.e. they appear right next to one another

*$1$ appears to the left of $2$
For $n=3$ the whole list is:  $123,312$
For $n=4$ the whole list is: $1234,1243,3124,4123,3412,4312$
For $n=5$ the list begins as: $12345,12354,12435,12453,\dots$ with several more yet unwritten.

To solve, apply multiplication principle to the following steps:


*

*Choose the location in the arrangement which the $1$ will occupy (remember that you must leave enough space to the right for the $2$ to occupy as well)

*Place the two adjacent to the right of the $1$

*From left-to-right fill in the remaining empty spaces in the arrangement with one of the remaining digits.
A: Some hints:


*

*How many places can you put $12$ in the sequence?

*How many ways can you arrange the other $n-2$ numbers in the spaces that remain?

