Let A = {1, 2, 3, ldots, 10} and B = left{fra | vedclass

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(A) We are given $A = \{1, 2, \ldots, 10\}$ and $B = \left\{\frac{m}{n} : m, n \in A, m < n, \gcd(m, n) = 1\right\}$.To find $n(B)$,we count the numbe

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$31$

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(A) We are given $A = \{1, 2, \ldots, 10\}$ and $B = \left\{\frac{m}{n} : m, n \in A, m To find $n(B)$,we count the number of fractions $\frac{m}{n}$ for each $n \in \{2, 3, \ldots, 10\}$ such that $m For $n=2$: $m \in \{1\}$,$\gcd(1, 2) = 1$. Count = $1$.For $n=3$: $m \in \{1, 2\}$,$\gcd(1, 3) = 1, \gcd(2, 3) = 1$. Count = $2$.For $n=4$: $m \in \{1, 3\}$,$\gcd(1, 4) = 1, \gcd(3, 4) = 1$. Count = $2$.For $n=5$: $m \in \{1, 2, 3, 4\}$,all are coprime to $5$. Count = $4$.For $n=6$: $m \in \{1, 5\}$,$\gcd(1, 6) = 1, \gcd(5, 6) = 1$. Count = $2$.For $n=7$: $m \in \{1, 2, 3, 4, 5, 6\}$,all are coprime to $7$. Count = $6$.For $n=8$: $m \in \{1, 3, 5, 7\}$,all are coprime to $8$. Count = $4$.For $n=9$: $m \in \{1, 2, 4, 5, 7, 8\}$,all are coprime to $9$. Count = $6$.For $n=10$: $m \in \{1, 3, 7, 9\}$,all are coprime to $10$. Count = $4$.Summing these counts: $1 + 2 + 2 + 4 + 2 + 6 + 4 + 6 + 4 = 31$.Thus,$n(B) = 31$.

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

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