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Let no of elements in each equivalence class be 'n'.

Now, asper handshaking theorem, there will be nP2 elemnts in the relation R corresponding to single equivalence class. Similarly , nPfor the remaining 2 classes.

So, total elements = (nP2)3

Here, in the options, only option (C) is proper cube. So, I think (C) is the answer.

Did not get the Handshaing Theorem in this context .. can u plz provide any link or resources on this part?

Handshaking means ordered pairs.

Sorry, it wont be nP2. It would be n2

e.g if the equivalence class is {1,2,3} then the ordered pairs will be n2 = 32 = 9

i.e R will contain ordered pairs

1,1

2,2

3,3

1,2

2,1

2,3

3,2

1,3

3,1

Right?

That's okay but did not get " total elements = (nP2)3 "

It should be (n2)3

why (n2)3 ?

did not understand the cube part..

There are 3 equivalence classes of same size. Thts why. Read the question carefully.

at least 3 equivalence class not exactly 3 is mentioned even if we consider 3 equivalence classes then total number of ordered pairs (according to your previous explanation) becomes 3*n^2 not (n2)3

Oops...I read the question wrong. But still consider

#total elements = k. n2

Now, try to find the factors of any of the answers and check if we can get a perfect square. Since,

k >= 3

we can take k=4 so that we get (D) as the answer.