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+2 votes

Choose the correct alternatives (More than one may be correct).

The number of rooted binary trees with $n$ nodes is,

  1. Equal to the number of ways of multiplying $(n+1)$ matriees.
  2. Equal to the number of ways of arranging $n$ out of $2 n$ distinct elements.
  3. Equal to $\frac{1}{(n+1)}\binom{2n}{n}$.
  4. Equal to $n!$.
asked in DS by Veteran (28.1k points)   | 46 views

1 Answer

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Number of BTs with Unlabelled nodes is (2n C n )/(n+1)

Number of BTs with labelled nodes is (2n C n ) * (n!) /(n+1) .


So, answer is option C.

If question have asked about BTs with labelled nodes then answer is B i.e picking n out of 2n people i.e 2n C n then n can be arranged in n! ways
answered by Active (2.3k points)  
still (n+1) term would be missing in option B rt?
yes arjun sir i think he forgot to mention but still they given " Number of BTs with labelled nodes is (2n C n ) * (n!) /(n+1)"
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