Given sets :

A={1,2,3,4, ............,10}

B={0,1,2,3,4}

We are looking for pairs $(a,b) \in A \times A$ such that :

$ 2(a-b)^2 + 3(a-b) \in B $

Let's break down the relation :

Case 1 : $ a-b = 0 $

$ 2(a-b)^2 + 3(a-b) = 0 $

Pairs : $(1,1), (2,2), (3,3), \ldots, (10,10)$ which gives 10 pairs.

Case 2 : $ a-b = 1 $

$ 2(a-b)^2 + 3(a-b) = 2(1) + 3(1) = 5 $

But 5 is not in B, so no pairs for this case.

Case 3 : $ a-b = -1 $

$ 2(a-b)^2 + 3(a-b) = 2(1) - 3(1) = -1 $

This value is not in B, so no pairs for this case.

Case 4 : $ a-b = 2 $

$ 2(a-b)^2 + 3(a-b) = 2(4) + 3(2) = 8+6 = 14 $

Again, 14 is not in B, so no pairs for this case.

Case 5 : $ a-b = -2 $

$ 2(a-b)^2 + 3(a-b) = 2(4) - 3(2) = 8 - 6 = 2 $

Pairs : $(1,3), (2,4), (3,5), (4,6), (5,7), (6,8), (7,9), (8,10)$ which gives 8 pairs.

For any other $ a-b $ value, the quadratic will grow larger than the maximum value in B, so we don't need to consider them.

In total, we have $ 10 + 8 = 18 $ pairs in the relation $ R $.

Therefore, the number of elements in the relation $ R $ is 18.