The number of permutations of 1, 2, 3, 4, 5, 6, 7, 8 and 9 taken all at a time such that digit 1 appearing somewhere to the left of 2, 3 appearing to the left of 4, and 5 somehere to the left of 6 is(eg 815723945 would be one such permutation) :–
| A. |
9.7! |
 |
| B. |
8! |
| C. |
5!4! |
| D. |
8!4! |
Number of digits are 9. Select two places for the digit 1 and 2 in `9C_2` ways. From the remaining seven places, select any two places for 3 and 4 in `7C_2` ways and from the remaining five places, select any two for 5 and 6 in `5C_2` ways. Now, the remaining three digits can be filled in 3! ways. Therefore,
