Question

Number of permutations of $1,2,3,4,5,6,7,8$ and $9$ taken all at a time, 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. $5!\cdot 4!$
• D. $8!\cdot 4!$

Answer 1

Answer: (A)

$9.7!$

There are $9!$ total permutation of the element of the set. ! is to the left of 2 in exactly half of these. 3 is also to the left of 4 in exactly half of the permutation, and 5 is to the left of 6 in exactly half of the permutation. These three events are totally independent of each other,

So the number we want to calculate$=\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot 9!=\frac{1}{8}\cdot 9\cdot 8\cdot 7!=9\cdot 7!$