Question

Number of permutations of $1,2,3,4,5,6,7,8$ and $9$ taken all at a time such that
$1$ appears to the left of $2$
$3$ appears to the left of $4$ and
$5$ appears to the left of $6$ is $k\times 7!$, then the value of $k$ is

Answer 1

Number of digits are $9$
Select $2$ places for $1$ and $2$ in ${}^9{C}_2$ ways,
Arrangement of them is fixed.
From the remaining $7$ places select any two places for $3$ and $4$ in ${}^7{C}_2$ ways
Arrangement of them is fixed.
Also, from the remaining $5$ places select any two for $5$ and $6$ in ${}^5{C}_2$ ways.
Arrangement of them is fixed.
Now, the remaining $3$ digits can be filled in $3!$ ways

$\therefore$ The total ways
$={}^9{C}_2\cdot {}^7{C}_2\cdot {}^5{C}_2\cdot 3!=\frac{9!}{2!.7!}\cdot \frac{7!}{2!.5!}\cdot \frac{5!}{2!.3!}\cdot 3!=\frac{9!}{8}=9\cdot 7!\therefore k=9$