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Consider 4 labeled 1,2,3,4. The number of distinct binary tree possible such that whose inorder traversal is 1,2,3,4 are ........
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it's  $\binom{2n}{n}$ / (n+1) = 14 right ??
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@Magma

yes, in each pattern we post the nodes as one way which give the inorder result as 1,2,3,4

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hmm thanks  Shaik Masthan

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Thats the thing i also did C(2n,n)/(n+1)  but in made easy they have also multiplied by n! extra.

BST are by default labelled tree right i dont know why they multiplied by n! there.
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If he did not strict to get in order $1,2,3,4$ then we can assign $4$ number in any order and get $4!$ combination,but in this question, it is not possible.

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