If A is a finite set containing 3 elements, the number of relations that can be defined on A is
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.A relation is simply a subset of cartesian product A×A.
If A×A=[(a
1
,a
1
),(a
1
,a
2
),.....(a
1
,a
n
),
(a
2
,a
1
),(a
2
,a
2
),........(a
2
,a
n
)
......
(a
n
,a
1
),(a
n
,a
2
).........(a
n
,a
n
)]
We can select first element of ordered pair in n ways and second element in n ways.
So, clearly this set of ordered pairs contain n
2
pairs.
Now, each of these n
2
ordered pairs can be present in the relation or can't be. So, there are 2 possibilities for each of the n
2
ordered pairs.
Thus, the total no. of relations is 2
n
2
.
Hence, option C is correct.
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