First time here? Checkout the FAQ!
+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 (32.9k points)   | 103 views

1 Answer

0 votes
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)"

Top Users Aug 2017

    4660 Points

  2. Bikram

    4366 Points

  3. akash.dinkar12

    3258 Points

  4. rahul sharma 5

    3042 Points

  5. manu00x

    2682 Points

  6. makhdoom ghaya

    2410 Points

  7. just_bhavana

    2100 Points

  8. Tesla!

    1918 Points

  9. stblue

    1682 Points

  10. joshi_nitish

    1608 Points

24,928 questions
32,024 answers
30,113 users