Let 𝑃 be the set of all integers from 1 to 15. Consider any order of…

To form a complete BST using numbers 1 to 15,
the final BST must be perfectly balanced.

Hence:

Root = 8

Its immediate children must be:

Left child = 4
Right child = 12

Now analyze insertion order.

1st inserted element:
Must be 8 (root)

2nd inserted element:
Must be either 4 or 12

3rd inserted element:
Must become the other child of root.

Therefore, the third inserted element can only be:

4 or 12

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Checking Options
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Option 1: 4
Possible as third insertion.

Option 2: 2
Possible if:
8, 4, 2
This can still later form a complete BST.

Option 3: 10
Possible if:
8, 12, 10

Option 4: 5
Impossible as third insertion.

Reason:
If 5 is inserted before 4 exists,
then 5 becomes direct left child of 8,
which violates the required complete BST structure.

Hence, 5 can NEVER be the third inserted element.

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Final Answer
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5
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