(C) is Correct -
Probability = $\frac{Number \ of \ Favorable\ Outcomes(min heaps)}{ Total \ Trees(13!) }$
Now, total Min-Heaps :
Firstly we select minimum element i.e. $1$ for root = 1 way
We have $3$ subtrees of $3,6,3$ sizes respectively
First Subtree with 3 elements:
Steps:
- Choose $3$ elements =$^{12}C_3$ ways
- Give minimum to root = 1 way
- $2$ elements , $2$ children(left & right) = 2 ways { Any of the $2$ elements can be given to left or right child}
Second Subtree with 3 elements :
Steps:
- Choose $3$ elements from $9$ left elements = $^{9}C_3$ ways
- Similar to above assign these = 2 ways
Subtree with 6 elements:{ We already have $6$ elements left now}
$1$ Root , $1$left_most child , $1$ right_most child , $3$ in middle **
Steps:
- Choose root = 1 way {minimum}
- Now $5$ elements left, choose leftmost child = 5 ways (choose anyone )
- Now $4$ elements left, choose right_most child = 4 ways (choose anyone )
- Now $3$ elements left for middle :
a) assign root = 1 way
b) assign left and right anyone = 2 ways
For, Total Heaps multiply all ways above.
Total probability = $\frac{ Total\ Heaps}{Total\ Trees(13!)}$
= $\frac{12! \ * \ 2\ *\ 9!\ * \ 2\ *\ 5\ * \ 4\ *\ 2}{13!\ *\ 3!\ *\ 9!\ *\ 3!\ *\ 6!}$
= $\left(\frac{1}{13}\right)\left(\frac{1}{6}\right)\left(\frac{1}{3}\right)^{3}$