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. $9\cdot 7!$
• B. $8!$
• C. $5!\cdot 4!$
• D. $8!\cdot 4!$

Answer 1

The correct option is A $9\cdot 7!$

There are exactly $9!$ ways in which we can arrange this set out of which 1 is to left of 2 in half of them out of which exactly half of them has 3 to left of 4 and from that exactly half of them has 5 to the left of 6. Hence,

answer = $(\frac{1}{2}{)}^3.9!=9.7!$