0

What is the number of relations on a n element set that are antisymmetric and not irreflexive? I have tried doing this as fallows- no of antisymmetric relations having atleast one self pair[like (x,x)]=3^((n^2-n)/2){nC1+nC2+.......+nCn}=3^((n^2-n)/2)(2^n-1). Is it correct?

Mayank Deora
  • 1,466
  • 1
  • 13
  • 26

1 Answers1

0

Since any relation on a set $A$ to itself can be represented by a Boolean matrix. Thus each relation corresponds to a $n \times n$ matrix (call it $M$).

For anti-symmetric relation you need the following: Let $i \neq j$ and let $m_{ij}$ be the $ij^{\text{th}}$ entry of $M$. Then

  1. If $m_{ij}=1$, then the entry $m_{ji}=0$.
  2. If $m_{ij}=0$, then $m_{ji}=1$ or $0$.
  3. The entries $m_{ii}=0$ or $1$.

With this in mind let us consider the pair of entries $(m_{ij}, m_{ji})$. For anti-symmetry these pairs could only be of the form $(1,0), (0,1)$ or $(0,0)$.

Number of such pairs (non-diagonal entry pairs) are $\dfrac{n^2-n}{2}$. Thus the number of matrices $M$ with such pairs are $3^{\frac{n^2-n}{2}}$. Now for antisymmtery the $n$ diagonal entries can be chosen $2^n$ ways (either $0$ or $1$). $$ \text{Thus the number of anti-symmteric relations is } 3^{\frac{n^2-n}{2}} \cdot 2^{n}. $$

For irreflexivity you require that at least one diagonal element of $M$ should be $1$. So we count for the opposite case when all diagonal entries are $0$. With the method illustrated above: the number of anti-symmteric relations with all diagonal entires zero (i.e. the relations which ARE irreflexive) is $3^{\frac{n^2-n}{2}}$.

Thus the number of anti-symmetric relations that are NOT irreflexive is $$ 3^{\frac{n^2-n}{2}} \cdot 2^{n} - 3^{\frac{n^2-n}{2}} = 3^{\frac{n^2-n}{2}}(2^n-1). $$

So your answer is correct!!.

Anurag A
  • 42,261