Here we have 11 nodes hence 11 positions. To satisfy topological ordering we have to fix node **a** at first position, node **k** at 11th position, and node **d** at 4th position.

Now we have 2,3,5,6,7,8,9,10 as empty positions. Among them we can fill positions 2 and 3 from nodes **b** or **c **only (hence **2** ways) and remaining 6 places can be filled by **e,g,i,f,h,j** such that relative ordering **e-g-i** and **f-h-j** must be maintained, this can be done in **6!/(3!3!) = 20** ways.

Total topological orderings = 2 x 20 =** 40**