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

<p>Given sets : <br/><br/>A={1,2,3,4, ............,10} <br/><br/> B={0,1,2,3,4} <br/><br/>We are looking for pairs $(a,b) \in A \times A$ such that : <br/><br/>$ 2(a-b)^2 + 3(a-b) \in B $</p> <p>Let&#39;s break down the relation :</p> <p><strong>Case 1 :</strong> $ a-b = 0 $ <br/><br/>$ 2(a-b)^2 + 3(a-b) = 0 $ <br/><br/>Pairs : $(1,1), (2,2), (3,3), \ldots, (10,10)$ which gives 10 pairs.</p> <p><strong>Case 2 :</strong> $ a-b = 1 $ <br/><br/>$ 2(a-b)^2 + 3(a-b) = 2(1) + 3(1) = 5 $ <br/><br/>But 5 is not in B, so no pairs for this case.</p> <p><strong>Case 3 :</strong> $ a-b = -1 $ <br/><br/>$ 2(a-b)^2 + 3(a-b) = 2(1) - 3(1) = -1 $ <br/><br/>This value is not in B, so no pairs for this case.</p> <p><strong>Case 4 :</strong> $ a-b = 2 $ <br/><br/>$ 2(a-b)^2 + 3(a-b) = 2(4) + 3(2) = 8+6 = 14 $ <br/><br/>Again, 14 is not in B, so no pairs for this case.</p> <p><strong>Case 5 :</strong> $ a-b = -2 $ <br/><br/>$ 2(a-b)^2 + 3(a-b) = 2(4) - 3(2) = 8 - 6 = 2 $ <br/><br/>Pairs : $(1,3), (2,4), (3,5), (4,6), (5,7), (6,8), (7,9), (8,10)$ which gives 8 pairs.</p> <p>For any other $ a-b $ value, the quadratic will grow larger than the maximum value in B, so we don&#39;t need to consider them.</p> <p>In total, we have $ 10 + 8 = 18 $ pairs in the relation $ R $.</p> <p>Therefore, the number of elements in the relation $ R $ is 18.</p>