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

 A = {1, 2, 3,......., 10}

$\mathrm{B}=\left\{\frac{\mathrm{m}}{\mathrm{n}}:\mathrm{m},\mathrm{n}\in \mathrm{A},\mathrm{m}<\mathrm{n}\text{ and }\gcd (\mathrm{m},\mathrm{n})=1\right\}$

$$\begin{gathered} \mathrm{n}\left(\mathrm{B}\right)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} \\ \left.\begin{array}{ll}\mathrm{n}=2& \left\{\frac{1}{2}\right\}\\ \mathrm{n}=3& \left\{\frac{1}{3},\frac{2}{3}\right\}\\ \mathrm{n}=4& \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left\{\frac{1}{4},\frac{3}{4}\right\}\\ \mathrm{n}=5& \left\{\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5}\right\}\\ \mathrm{n}=6& \left\{\frac{1}{6},\frac{5}{6}\right\}\\ \mathrm{n}=7& \left\{\frac{1}{7},\frac{2}{7},\frac{3}{7},\frac{4}{7},\frac{5}{7},\frac{6}{7}\right\}\\ \mathrm{n}=8& \left\{\frac{1}{8},\frac{3}{8},\frac{5}{8},\frac{7}{8}\right\}\\ \mathrm{n}=9& \left\{\frac{1}{9},\frac{2}{9},\frac{4}{9},\frac{5}{9},\frac{7}{9},\frac{8}{9}\right\}\\ \mathrm{n}=10& \left\{\frac{1}{10},\frac{3}{10},\frac{7}{10},\frac{9}{10}\right\}\end{array}\right\} total 31 \end{gathered}$$