Question

Number of permutations of 1, 2, 3, 4, 5, 6, 7, 8 and 9 taken all at a time are such that the digit 1 appearing somewhere to the left of 2, 3 appearing to the left of 4 and 5 somewhere to the left of 6, is (e.g. 815723946 would be one such permutation)

Answer 1

There are $9!$ total permutations of the elements of that set. $1$ is to the left of $2$ in exactly half of these. $3$ is also to the left of $4$ in exactly half of the permutation. These three events are totally independent to each other, So, the number we want to calculate is $\frac{1}{2},\times \frac{1}{2},\times \frac{1}{2},\times 9!=\frac{1}{8}\times 9\times 8\times 7!=9\times 7!$