Question

Number of permutations of 1,2,3,4,5,6,7,8 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

Answer: (A)

9.7!

There are $91$ permutations possible of the set [$1,2,3,4,5,6,7,8,9$}

$1$ is to the left of $2$ in exactly half of these.

$3$ is to the left of $4$ in exactly half of the permutations and similarly half of the permutations contain $5$ to the left of $6$. There by the total permutations for the above three independent cases is $\frac{1}{2}.\frac{1}{2}.\frac{1}{2}.91=\frac{1}{8}.9.8.7!=9\times 7!$

Alternatively,

Consider the arrangement {$1,2$}, we can choose $9_{C_2}$ places for such an arrangement. Then $7_{C_2}$ places for {$3,4$} and then $5_{C_2}$ places for {$s,b$} and then in $3!$ ways we can arrange $7,8$ & $9$ giving used total of

$9_{C_2}\times 7_{C_2}\times 5_{C_2}\times 3!=9\times 7!$

Note : {$1,2$} does not mean that $1,2$ are together. It just implies that $1$ is to the left of $2$.