Let $$A = 1, 2, 3, 4, \ldots, 10$$ and $$B = 0, 1, 2, 3, 4$$. The number of elements in the relation $$R = \{(a, b) \in A \times A: 2a - b^2 + 3a - b \in B\}$$ is ________.

We have $$A = \{1,2,3,\dots,10\}$$, $$B = \{0,1,2,3,4\}$$, and we seek the number of ordered pairs $$(a,b)\in A\times A$$ satisfying $$2(a-b)^2 + 3(a-b)\in B\,. $$ To analyze this, set $$d = a - b$$ and define $$f(d) = 2d^2 + 3d\,. $$ We look for integer values of $$d$$ such that $$f(d)\in B\,. $$

When $$d = 0\,,\;f(0) = 0\in B$$, which accounts for the 10 diagonal pairs $$(a,a)\,,\;a=1,2,\dots,10\,. $$ For $$d = 1\,,\;f(1) = 5\notin B$$, and for $$d = -1\,,\;f(-1) = 2 - 3 = -1\notin B\,. $$ When $$d = 2\,,\;f(2) = 8 + 6 = 14\notin B$$, but for $$d = -2\,,\;f(-2) = 8 - 6 = 2\in B$$. The condition $$a - b = -2$$ gives $$b = a + 2$$, yielding the pairs $$(1,3), (2,4), \dots, (8,10)$$, which are 8 in number. All other values of $$d$$ produce $$f(d)$$ outside of $$B\,. $$

In total there are $$10 + 8 = 18$$ ordered pairs satisfying the condition. The answer is 18.