Hi.. :)

We need to find no. of permutations such that numbers 1 and 2 always precede 3 and 6 and 7 always succeed the number 3.

It means "3" can come only at "third", "fourth" , "fifth" place.

Now, fix "3" at "third" position.

Number of ways would be : 2!*4!

Now, fix "3" at "fourth" position.

Number of ways would be : 3!*3!

Now, fix "3" at "fifth" position.

Number of ways would be 4!*2!.

Just add all of them to get total number of ways.

:)