Basically root is fixed for 1 and right of root is 7. Since right of root is 7 then 2 should come as left node of root and 3 as any child of 2 , nine is child of 3 .so our tree will look like this
Now 3 and 9 means root 3 and child 9 has 4 places to move
so we are basically left with 8,10,11,12,13,14,15, these can come below 7 so selecting 6 out of 7 can be done in C(7,6) as root is fixed as 7, these 7 node subtree min heap can be made in 80 ways
so we have 4*C(7,6)*80
Now we have 4 places left in left subtree below 2. So we can select 1 in C(4,1) and permute them in 2 ways.
so Total ways: 4*C(7,6)*80*C(4,1)*2 ways
I understood till here, but confused why we are not permuting in 3 ways for last multiplication.