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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 $n!$.
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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) .

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
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)"