How many relations on set {a,b,c} are reflexive and antisymmetric? Question 1. I have seen the matrix way to solve this question, I am wondering how to use combinatorics in calculating the number of antisymmetric relations and then throw in reflexive tuples. 
My idea as following, $3$ elements in a set which will generate $9$ tuples, $3$ of which are "reflexive tuples" and the rest will be in pairs as symmetric. 
$2^6$ is the total number of a reflexive relation, then minus not antisymmetric relations. {a,b,c} are obviously distinct, if both "symmetric pairs in the reflexive relation, then it's not antisymmetric" Then it turns out $2^6 -2^3 =56$. The answer should be $27$. I don't see what has gone wrong here.
Please note that I didn't claim the complement of symmetric is antisymmetric, but under the constraint of reflexive relation it is.
Question 2. Asking for further clarification, if symmetric and antisymmetric implies reflexive? 
Question 3. If an empty set or any set with one tuple only could claim that it's antisymmetric?
Thanks for any help
 A: In this answer $a,b,c$ are distinct.
For the flaw in your reasoning see the comment of Brian.
Let us count the number of relations on $\{a,b,c\}$ that are reflexive and are not anti-symmetric.
We can write the set of relations as: $$\mathcal R(\{a,b\})\cup\mathcal R(\{a,c\})\cup \mathcal R(\{b,c\})$$where e.g. $R\in\mathcal R(\{a,b\})$ if and only if $R$ is reflexive relation on $\{a,b,c\}$ with $\langle a,b\rangle,\langle b,a\rangle\in R$.
Applying inclusion/exclusion and symmetry we find:$$|\mathcal R(\{a,b\})\cup\mathcal R(\{a,c\})\cup \mathcal R(\{b,c\})|=$$$$3|\mathcal R(\{a,b\})|-3|\mathcal R(\{a,b\})\cap\mathcal R(\{a,c\})|+|\mathcal R(\{a,b\})\cap\mathcal R(\{a,c\})\cap \mathcal R(\{b,c\})|=$$$$3\cdot2^4-3\cdot2^2+1=37$$
So the number of relations on $\{a,b,c\}$ that are reflexive and anti-symmetric equals:$$2^6-37=64-37=27$$
question 2: 
The empty relation on any set is symmetric and anti-symmetric but is only reflexive if it is looked at as a relation on the empty set.
question 3: (I am not sure whether I understand your question correctly)
If a set is empty then there is only one relation on that set, which is the empty relation. The empty relation is anti-symmetric (as said above).
If a set is a singleton $\{a\}$ then there are two relations on it: the empty relation and the relation $\{\langle a,a\rangle\}$. Both are anti-symmetric.
A: A relation $R$ is symmetric over set $S$ when $\forall x\in S~.(x,x)\in R$.  
The relation is antisymmetric when $\forall x\in S~\forall y\in S~.((x,y)\in R\wedge (y,x)\in R\to x=y)$
So for any $x>y$ either $[(x,y)\in R\wedge(y,x)\notin R]$, $[(x,y)\notin R\wedge(y,x)\in R]$, or $[(x,y)\notin R\wedge(y,x)\notin R]$ will be the case.  That is for every pair $(x,y)$ where $x>y$ there will be three choices (include the pair, include its inverse, or include neither).
So the count of relations over $\{a,b,c\}$ which will be reflexive and antisymmetric will be the count of ways to make three independent choices with three options.


Question 2. Asking for further clarification, if symmetric and antisymmetric implies reflexive?

No.  Neither of symmetry nor antisymmetry alone implies reflexivity.  Together they both imply that if any elements are related, then they will be the same element; that is not the same thing as reflexivity.  $$\forall x\in S~\forall y\in S~.((x,y)\in R\to x=y)$$

Question 3. If an empty set or any set with one tuple only could claim that it's antisymmetric?

A relation is not antisymmetric if there exists any pair of non-identical members and its inverse that are both in the relation.  $$\exists x\in S~\exists y\in S~.(x\neq y\wedge (x,y)\in R\wedge(y,x)\in R)$$
So if there exists none such, the relation is antisymmetric.
