Question

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

• A. 9.7!
• B. 8!
• C. 5!.4!
• D. 8!-4!

Answer 1

The correct option is A 9.7!

Total no. of permutations $=9!$

In these, half have $1$ in left of $2,$

half have $3$ in left of $4,$

half have $5$ in left of $6,$

$\therefore$ No. of required permutations $=\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times 9!$

$=9\times 7!$

Hence, the answer is $9\times 7!.$