Heap with $15$ elements will be a **complete binary tree**, hence it will have $8$ leaf nodes, and we want all the leaf nodes to be greater than all the non-leaf nodes, hence we can only have largest $8$ elements in leaf nodes. If we place any other elements in leaf nodes then there will be at least one of the largest $8$ elements which we will need to adjust in a non-leaf node, hence given condition will be violated. For example, if the elements are integers from $1$ to $15$ then the elements $[8,15]$ must be present in the leaf nodes, if we place any other element (which will be definitely $<8$) in any leaf node, then one element from $[8,15]$ needs to be adjusted in a non-leaf node, and this adjusted non-leaf node will have element $>$ the element that we placed in the leaf node.

Also, we need the minimum element at the root node,hence the structure of the heap will be:

** minimum**

** / \**

** x x **

** / \ / \**

** x x x x**

** / \ / \ / \ / \**

**o o o o o o o o**

Here **o **denotes a leaf node, **x **denotes a non-leaf node.

Now, for the $8$ leaf nodes, we can place the $8$ largest elements in any order, and there are $8!$ such orders **…………. (i)**

So, we have placed $8$ leaf nodes and the root node, now we are left with $15-8-1=6$ elements.Now, the nodes that we need to fill are represented by **x **in the above heap. From the structure, it is very clear that $3$ of the remaining $6$ elements will be on the left side and remaining $3$ will be on the right side. So, we can select any $3$ of the remaining $6$ elements to be placed in the left sub tree, total ways to do this will be $^6C_3$. **…………...(ii)**

Now, for the selected three elements, we need to ensure that the parent element is smaller than both the children, i.e. in the following structure:

a

/ \

b c

we want to make sure that $a<b$ and $a<c$ to satisfy the property of min-heap, hence, $a$ must be the smallest element among the $3$ elements, and we can also interchange $b$ and $c$ if the above conditions are satisfied, because we know that the leaf nodes are already greater than all the non-leaf nodes. Hence, there are $2$ ways to fill the left part, i.e. we can keep $(b,c)$ or $(c,b)$ **………...(iii)**

Similarly, there are $2$ ways to fill right part. ** .…….……...(iv)**

Hence, total number of ways will be: $8!*^6C_3*2*2=3225600$