Let $A=\{1,2,3, \ldots, 10\}$ and $B=\left\{\frac{m}{n}: m, n \in A, m< n\right.$ and $\left.\operatorname{gcd}(m, n)=1\right\}$. Then $n(B)$ is equal to :

<p>To find the number of elements in set $ B$, we consider pairs $\left(\frac{m}{n}\right)$ where $ m, n \in A $ with $ m < n $ and $\text{gcd}(m, n) = 1$.</p> <p>Here's the breakdown for each possible $ m $:</p> <p><p><strong>For $ m = 1 $:</strong> </p></p> <p><p>Possible values for $ n $ are $ 2, 3, 4, 5, 6, 7, 8, 9, 10 $. </p></p> <p><p>Total pairs: $ 9 $.</p></p> <p><p><strong>For $ m = 2 $:</strong> </p></p> <p><p>Possible values for $ n $ are $ 3, 5, 7, 9 $ (since these have $\text{gcd}(2, n) = 1$). </p></p> <p><p>Total pairs: $ 4 $.</p></p> <p><p><strong>For $ m = 3 $:</strong> </p></p> <p><p>Possible values for $ n $ are $ 4, 5, 7, 8, 10 $. </p></p> <p><p>Total pairs: $ 5 $.</p></p> <p><p><strong>For $ m = 4 $:</strong> </p></p> <p><p>Possible values for $ n $ are $ 5, 7, 9 $. </p></p> <p><p>Total pairs: $ 3 $.</p></p> <p><p><strong>For $ m = 5 $:</strong> </p></p> <p><p>Possible values for $ n $ are $ 6, 7, 8, 9 $. </p></p> <p><p>Total pairs: $ 4 $.</p></p> <p><p><strong>For $ m = 6 $:</strong> </p></p> <p><p>Possible value for $ n $ is $ 7 $. </p></p> <p><p>Total pairs: $ 1 $.</p></p> <p><p><strong>For $ m = 7 $:</strong> </p></p> <p><p>Possible values for $ n $ are $ 8, 9, 10 $. </p></p> <p><p>Total pairs: $ 3 $.</p></p> <p><p><strong>For $ m = 8 $:</strong> </p></p> <p><p>Possible value for $ n $ is $ 9 $. </p></p> <p><p>Total pairs: $ 1 $.</p></p> <p><p><strong>For $ m = 9 $:</strong> </p></p> <p><p>Possible value for $ n $ is $ 10 $. </p></p> <p><p>Total pairs: $ 1 $.</p></p> <p>Adding all these up, the total number of elements in set $ B $ is:</p> <p>$ 9 + 4 + 5 + 3 + 4 + 1 + 3 + 1 + 1 = 31 $</p>