Let us take 15 elements [1,2,3,4.....15]. To satisfy the condition mentioned in the problem all the 8 largest element should come to the last level. means element {8,9,10,11,12,13,14,15} must come in the last level.
So total no of ways to arrange the elements in the last level = 8!
now for the remaining elements {1,2,3,4,5,6,7} there are no restriction hence they can come in any order in min heap. Hence it will be equivalent to construct the min heap using elements {1,2,3,4,5,6,7}. = 80 ways. for this please see below. link http://karmaandcoding.blogspot.com/2012/02/number-of-min-heaps-from-array-of-size.html
hence total no will be = 80 * 8! = 3225600.
If the same question is little bit modified and let us say - The number of min heap trees are possible with 15 elements such that every node at a particular level must be greater than all the nodes of the tree which are above that level. ex. all the nodes present at level 3 must be greater than all the nodes present at level 2 and so on....
Then the explanation given by @Aman Chauhan seems more meaningful.
Suppose consider 15 elements 1,2,3,4,....15.
It is min heap ,level by level elements are stored.Root is at 1st level.
1st level; 1
2nd level: 2,3
3rd level: 4,5,6,7
4th level: 8,9,10,11,12,13,14,15.
In the second level elements are nodes 2,3 occupies 2 ways.no matter because it satisfies the heap property.
In the 3rd level also same nodes are 4,5,6,7 can be arranged in 4! ways.
In the 4th level 8!
SO TOTAL NO OF TREES ARE 1!*2!*4!*8*=1935360