Question

Number of permutations 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)

(a) 9 . 7!

(b) 8!

(c) 5! . 4!

(d) 8! . 4!

Answer 1

Correct option (a)

Explanation:

Number of digits are 9

Select 2 places for the digit 1 and 2 in 9C2 ways from the remaining 7 places select any two places for 3 and 4 in 7C2 ways and from the remaining 5 places select any two for 5 and 6 in 5C2 ways Now, the remaining 3 digits can be filled in 3! ways

∴Total ways = 9C2 . 7C2 . 5C2 . 3!