Tedious permutation

There are $9!$ possible permutations of $\{1,2,3,4,5,6,7,8,9\}$ :

1. There are $\frac{9!}{2!}$ possible permutations with 1 coming before 2.

2.- There are $\frac{9!}{2!\cdot2!}$ possible permutations with 1 coming before 2, and 3 coming before 4. And finally,

3.- There are $\frac{9!}{2!\cdot2!\cdot2!} = 45360$ possible permutations with 1 coming before 2, 3 coming before 4 and 5 coming before 6

First we select 6 positions out of the 9 positions for the digits 1,2,3,4,5 and 6. This can be done in $\binom{9}{6}$ ways. For each position, there are $\binom{6}{2}$ ways for the 1 comes before 2, there are $\binom{4}{2}$ ways for the 3 comes before 4 and $\binom{2}{2}$ ways for the 5 comes before 6.We can arrange the remaining digits (7,8 and 9) in 3! ways. Thus, there are $\binom{9}{6}$ * $\binom{6}{2}$ * $\binom{4}{2}$ * $\binom{2}{2}$ * 3! =45360 such permutations.