The answer is 20.
See below image
OR
One way to select root.
10
/ \
LST (3 elements) RST (2 elements)
Now LST contains an element out of 5 (As the root is chosen)
So Total ways to choose element $_{}^{5}\textrm{C}_{3}$
and RST contains 2 element So number of ways = $_{}^{2}\textrm{C}_{2}$
Let LST chosen { 20,30,40}
Now 2 chosen as root of LST (in one way)
Now remaining 2 element can arrange in 2!
ie. 20 20
/ \ / \
30 40 40 30
Therefore Total Possibility = $_{}^{5}\textrm{C}_{3}$ x 2! x $_{}^{2}\textrm{C}_{2}$
= 20