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What is the possible number of reflexive relations on a set of 5 elements?

  1. $2^{10}$
  2. $2^{15}$
  3. $2^{20}$
  4. $2^{25}$
asked in Set Theory & Algebra by Boss (18.2k points)
edited by | 1.7k views
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2 Answers

+26 votes
Best answer

A relation consists of set of ordered pairs $(a,b)$. Here $a$ can be chosen in $n$ ways and similarly $b$ can be chosen in $n$ ways. So, totally $n^2$ possible ordered pairs are possible for a relation. Now each of these ordered pair can either be present in the relation or not- 2 possibilities for each of the $n^2$ pair. So, total number of possible relations =  $$2^{\left(n^2\right)}$$

Now, for a relation $R$ to be reflexive, ordered pairs $\left\{(a,a) \mid a \in S \right\}$, must be present in $R$. i.e.; the relation set $R$ must have $n$ ordered pairs fixed. So, number of ordered pairs possible is $ n^2 - n$ and hence total number of reflexive relations is equal to $$ 2^{\left(n^2-n\right)}$$

for $n=5$, answer will be, $2^{5^2-5}=2^{20}$

therefore $option\ C$ is correct

answered by Veteran (358k points)
edited by
+7 votes
The number of reflexive relations =2^(n^2 - n) so option c is correct
answered by Boss (14.3k points)


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