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 $\text{gcd}(m, n) = 1$.

Here's the breakdown for each possible $ m $:

For $ m = 1 $:

Possible values for $ n $ are $ 2, 3, 4, 5, 6, 7, 8, 9, 10 $.

Total pairs: $ 9 $.

For $ m = 2 $:

Possible values for $ n $ are $ 3, 5, 7, 9 $ (since these have $\text{gcd}(2, n) = 1$).

Total pairs: $ 4 $.

For $ m = 3 $:

Possible values for $ n $ are $ 4, 5, 7, 8, 10 $.

Total pairs: $ 5 $.

For $ m = 4 $:

Possible values for $ n $ are $ 5, 7, 9 $.

Total pairs: $ 3 $.

For $ m = 5 $:

Possible values for $ n $ are $ 6, 7, 8, 9 $.

Total pairs: $ 4 $.

For $ m = 6 $:

Possible value for $ n $ is $ 7 $.

Total pairs: $ 1 $.

For $ m = 7 $:

Possible values for $ n $ are $ 8, 9, 10 $.

Total pairs: $ 3 $.

For $ m = 8 $:

Possible value for $ n $ is $ 9 $.

Total pairs: $ 1 $.

For $ m = 9 $:

Possible value for $ n $ is $ 10 $.

Total pairs: $ 1 $.

Adding all these up, the total number of elements in set $ B $ is:

$ 9 + 4 + 5 + 3 + 4 + 1 + 3 + 1 + 1 = 31 $