in Combinatory edited by
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Match the pairs in the following questions by writing the corresponding letters only.
$$\begin{array}{|c|l|c|l|} \hline A. & \text{The number of distinct binary tree} & P. & \frac{n!}{2} \\& \text{ with n nodes.}\\ \hline B. &  \text{The number of binary strings of the length } & Q. & \binom{3n}{n} \\& \text{of 2n with an equal number of 0’s and 1’s}  \\ \hline C. & \text{The number of even permutation of n } & R. & \binom{2n}{n} \\& \text{ objects.}\\ \hline D. & \text {The number of binary strings of length 6n } & S. & \frac{1}{1+n}\binom{2n}{n} \\& \text{which are palindromes with 2n 0’s.} \\ \hline \end{array}$$
in Combinatory edited by
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4 Comments

The editing of this question has gone haywire. Some part of B and D is showing up in Q and S section. Same is the case in GO hardcopy.
4
true
0

Hey Folks,

Quick FYI, This question is not correctly formatted in the Gate overflow PDF.

4
Lakshman Patel RJIT

Thank you for fixing this question
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4 Answers

37 votes
 
Best answer
  1. $- S$  Catalan number https://gatecse.in/number-of-binary-trees-possible-with-n-nodes/
  2. $- R$  Choosing $n$ locations for $0$'s out of $2n$ locations. The remaining $n$ locations are filled with $1$'s (no selection required).
  3. $- P$  An even permutation is a permutation obtainable from an even number of two-element swaps, For a set of $n$ elements and $n>2$, there are $n!/2$ even permutations.
    Ref - http://mathworld.wolfram.com/EvenPermutation.html
  4.  $- Q$ 

Length $= 6n$, as it is palindrome, we need to select only the first half part of the string.

Total length to consider is $3n$ (Remaining $3n$ will be revese of this $3n$)

Now, choose $n \ 0's$ out of $3n$. So Q is correct for D.

edited by

8 Comments

Explain even permutations with an example
0
What is even permutation ???I'm not getting from above link ??
0

The option for the number of binary trees is incorrect. 

Number of BSTs (Binary Search Trees) = nth Catalan Number, whereas number of Binary Tress = (n!) * nth Catalan number.

Source : https://www.geeksforgeeks.org/total-number-of-possible-binary-search-trees-with-n-keys/

1
0
then what will be the value of odd permutations?
0

same as even permutations, $\frac{n!}{2}$

2
Number of Binary trees possible with n nodes:

https://gatecse.in/number-of-binary-trees-possible-with-n-nodes/
0
3 votes
Answer:

A - S
B - R
C - P
D - Q
3 votes

It will be $\frac {n!}{2}$

using the property of symmetry.

http://mathworld.wolfram.com/EvenPermutation.html

3 votes

An even permutation is a permutation obtainable from an even number of two-element swaps, For initial set {1,2,3,4}, the twelve even permutations are those with zero swaps: ({1,2,3,4}); and those with two swaps: ({1,3,4,2}, {1,4,2,3}, {2,1,4,3}, {2,3,1,4}, {2,4,3,1}, {3,1,2,4}, {3,2,4,1}, {3,4,1,2}, {4,1,3,2}, {4,2,1,3}, {4,3,2,1}). etc.

For a set of  n elements and n>2, there are n! / 2 even permutations, which is the same as the number of odd permutations

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