Answer / hitesh viradiya

m = 1 2 3 4 5 6 7 8
n=1 1
2 1 1
3 2 2 1
4 5 5 3 1
5 14 14 9 4 1
6 42 42 28 14 5 1
7 132 132 90 48 20 6 1
8 429 429 297 165 75 27 7 1

(2n)!/[(n+1)!n!]

Answer / kathir

There are 2 pointers available for each node.
So we can have 2*n pointers totally.

Total no. of edges = n-1

So, Null pointers = n+1.

We need to choose (n-1) pointers from 2n pointers.

So the combination results in (2n)C(n-1).

We can have n distinct roots possible.

So, answer will be (2n) C (n-1) / n.

which leads to,

{2n C n}/{n+1}. ( Unlabelled )


Labelled Structured tree will be,

{2n C n}/{n+1} * {n!}

Answer / durga prasad sahoo

No. of labeled binary tree :

n^(n-2)

No. unlabeled binary tree :

(2n)!/[(n+1)!.n!] (this is known as catlon number)

Answer / atul barve

No. of labeled binary tree :

{(2n)!/[(n+1)!.n!]}*n!

No. unlabeled binary tree :

(2n)!/[(n+1)!.n!] (this is known as catlon number)

Answer / neha shah

(2n)!/(n+1)!*n!

Answer / fawad ghafoor

The number of different trees with 8 nodes is 248
2^n - n

Answer / ajeet

int countTrees(int num)
{
if(num<=1)
return 1;
else
{
int root,left,right,sum=0;
for(root=1;root<=num;root++)
{
left=countTrees(root-1);
right=countTrees(num-root);
sum+=left*right;
}
return sum;
}
}

Answer / munyazikwiye pierre celestin

(2n)!/[(n+1)!*n!]

Answer / sayandip ghosh

The max number of binary trees that can be formed from n
nodes is given by the Catlan Number C(n).

C(n) = (2n)! / (n+1)!*n! for n>=0.

int findNoTree(int low ,int high)
{
int sum=0;
if(low<=high)
{
for(int k=low;k<=high;k++)
{
if(k==low)
sum+=findNoTree(low+1,high);
else
{
if(k==high)
sum+=findNoTree(low,high-1);
else

sum=sum+findNoTree(low,k-1)*findNoTree(k+1,high);
}
}

return sum;
}
return 1;
}

Answer / sachin

(2n)!/(n+1)!

main( ) { void *vp; char ch = ‘g’, *cp = “goofy”; int j = 20; vp = &ch; printf(“%c”, *(char *)vp); vp = &j; printf(“%d”,*(int *)vp); vp = cp; printf(“%s”,(char *)vp + 3); }

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