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

Options

A. `9.7!`
B. `8!`
C. `5!4!`
D. `8!4!`

Answer

A

Explanation

`(a)` Number of digits are `9`
Select `2` places for the digit `1` and `2` in `"^(9)C_(2)` ways from the remaining `7` places, select any two places for `3` and `4` in `"^(7)C_(2)` ways and from the remaining `5` places select any two for `5` and `6` in `"^(5)C_(2)` ways.
`:.` Total ways `="^(9)C_(2)*^(7)C_(2)*^(5)C_(2)*3!`
`=(9!)/(2!7!)*(7!)/(2!5!)*(5!)/(2!3!)*3!`
`=(9!)/(8)=(9*8*7!)/(8)=9*7!`