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Consider the following graph:

Find the total number of max-heap possible orderings with elements 12, 10, 1, 5, 7, 9, 8 such that each element is filled in one node of the above tree and element 10 occupies only the left child node of its parent.

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Except this we don’t have any other option to put 12 and 10 with fo

Now for childs of 10 we have   $\binom{5}{2}$ option and we can arrange them in $2!$ ways. Total=$\binom{5}{2}$*$2!$=20

Now on right side we don’t have any option because of left screw stree.

So Answer=20.

Answer:

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