Hello Aakash!

I am giving the solution of your query below:-

We have, X= {a, b, c, d}

The no. of distinct possible partitions= 4C1+ 4C2+ 4C3+ 4C4 = 4+ (4×3)/2+ 4+ 1 = 15

So, the no. of distinct partitions is 15.

The partitions are::-

{a}, {b}, {c},{d}, {a,b}, {b,c}, {c,d}, {a,d},{a,c}, {b,d}, {a,b,c}, {a,b,d}, {a,b,c,d} .

Equivalence relation are those which satisfies all three relation refexive,transitive and symmetric.

These are the equivalence relations formed on X:-

(a,b), (b,c), (c,a), (a,a), (b,b), (c,c), (d,d), (a,b), (d,a),(b,d), (d,b), (d,c), (c,d), (a,b,c), (a,b,d), (a,c,d), (b,c,d),(a,b,c,d)

Hope you are clear now!