Would some be able to check my work: for a set, how many different relations are there? I'm not too sure about the work that I have done here and would love if someone would be able to check my work. If it's correct, would you be able to explain the reason it works? (I feel as though I'm memorizing how to do these rather than understanding perfectly). If not correct, mind showing me the correct way?
For a set $$A=\{m,n,o,p,e,d,a,c,b,1,2,3,4,5\}$$

A) How many different relations one can define on the set?

I said $2^{14^2} = 2^{196}$.

B) How many different relations $T$ are there on $A$ where $(m,n)\in T$, $(3,4)\in T$, and $(a,c)\in T$?

I said $2^{(14^2)-3} = 2^{193}$

C) How many different symmetric relations are there?

I said $2^{\textstyle\binom{15}{2}} = 2^{105}$
Thank you in advance!
 A: Your answer to (A) is correct. 
A relation on $A$ is a subset of $A\times A$. $|A|=14$, so $|A\times A|=14^2=196$. Finally, a set $S$ with $n$ elements has $2^n$ subsets. To build a subset of $S$, you must decide for each of the $n$ elements $s$ of $S$ whether $s$ is in the subset or not; that’s a total of $n$ two-way choices, so they can be made in altogether $2^n$ different ways. In particular, $A\times A$, being a set of $196$ elements, has $2^{196}$ subsets.
Your answer to (B) is also correct. Once you decide that the three ordered pairs $\langle m,n\rangle,\langle 3,4\rangle$, and $\langle a,c\rangle$ are in $T$, you still have $14^2-3=193$ ordered pairs that you can choose to put into $T$ or leave out. In other words, to form $T$ you must choose some subset of the $193$ remaining ordered pairs, and there are $2^{193}$ such subsets.
(C) And again your answer is correct.
If $R$ is a symmetric relation on $A$, then we know that for any $x,y\in A$, either $\langle x,y\rangle$ and $\langle y,x\rangle$ are both in $R$, or neither is in $R$. There are $\binom{14}2$ (unordered) pairs of distinct elements of $A$. If $\{x,y\}$ is one of these pairs, we can either put both $\langle x,y\rangle$ and $\langle y,x\rangle$ into $R$ or leave both out; that’s one decision for the pair. Thus, there are $2^{\binom{14}2}$ possible sets of pairs that we can use. These give us the totally irreflexive symmetric relations on $A$, the ones that contain none of the ordered pairs of the form $\langle x,x\rangle$. We can combine any of these $2^{\binom{14}2}$ relations with any set of ordered pairs of the form $\langle x,x\rangle$ and still have a symmetric relation. Since $|A|=14$, there are $14$ such pairs and therefore $2^{14}$ sets of them. That gives us a total of 
$$2^{\binom{14}2}\cdot 2^{14}=2^{\binom{14}2+14}$$
symmetric relations on $A$. Finally,
$$\binom{14}2+14=\frac{14\cdot13}2+14=\frac{14\cdot13+2\cdot14}2=\frac{15\cdot14}2=\binom{15}2\;,$$
as in your answer. More generally,
$$\binom{n}2+n=\frac{n(n-1)}2+n=\frac{n(n-1)+2n}2=\frac{n(n+1)}2=\binom{n+1}2\;.$$
