116757 205 students take an examination of whom 105 pass in English, 70 students pass in mathematics and 30 students pass in both. How many students in both subjects?

116758 If the total number of \(m\)-element subsets of the set \(A=\left\{a_1, a_2, \ldots, a_n\right\}\) is \(k\) times the number of \(m\) element subsets containing \(a_4\), then \(n\) is

116759 In a statistical investigation of 1003 families of Calcutta, it was found that 63 families has neither a radio nor a T.V, 794 families has a radio and 187 has T.V. The number of families in that group having both a radio and \(a\) T.V is

116760 Let \(A, B, C\) be finite sets, Suppose that \(n(A)=\) \(10, \mathbf{n}(\mathrm{B})=15, \mathrm{n}(\mathrm{C})=20, \mathrm{n}(\mathrm{A} \cap \mathrm{B})=8\) and \(n(B \cap C)=9\). Then the possible value of \(\mathbf{n}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})\) is

116757 205 students take an examination of whom 105 pass in English, 70 students pass in mathematics and 30 students pass in both. How many students in both subjects?

116758 If the total number of \(m\)-element subsets of the set \(A=\left\{a_1, a_2, \ldots, a_n\right\}\) is \(k\) times the number of \(m\) element subsets containing \(a_4\), then \(n\) is

116759 In a statistical investigation of 1003 families of Calcutta, it was found that 63 families has neither a radio nor a T.V, 794 families has a radio and 187 has T.V. The number of families in that group having both a radio and \(a\) T.V is

116760 Let \(A, B, C\) be finite sets, Suppose that \(n(A)=\) \(10, \mathbf{n}(\mathrm{B})=15, \mathrm{n}(\mathrm{C})=20, \mathrm{n}(\mathrm{A} \cap \mathrm{B})=8\) and \(n(B \cap C)=9\). Then the possible value of \(\mathbf{n}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})\) is

116757 205 students take an examination of whom 105 pass in English, 70 students pass in mathematics and 30 students pass in both. How many students in both subjects?

116758 If the total number of \(m\)-element subsets of the set \(A=\left\{a_1, a_2, \ldots, a_n\right\}\) is \(k\) times the number of \(m\) element subsets containing \(a_4\), then \(n\) is

116759 In a statistical investigation of 1003 families of Calcutta, it was found that 63 families has neither a radio nor a T.V, 794 families has a radio and 187 has T.V. The number of families in that group having both a radio and \(a\) T.V is

116760 Let \(A, B, C\) be finite sets, Suppose that \(n(A)=\) \(10, \mathbf{n}(\mathrm{B})=15, \mathrm{n}(\mathrm{C})=20, \mathrm{n}(\mathrm{A} \cap \mathrm{B})=8\) and \(n(B \cap C)=9\). Then the possible value of \(\mathbf{n}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})\) is

116757 205 students take an examination of whom 105 pass in English, 70 students pass in mathematics and 30 students pass in both. How many students in both subjects?

116758 If the total number of \(m\)-element subsets of the set \(A=\left\{a_1, a_2, \ldots, a_n\right\}\) is \(k\) times the number of \(m\) element subsets containing \(a_4\), then \(n\) is

116759 In a statistical investigation of 1003 families of Calcutta, it was found that 63 families has neither a radio nor a T.V, 794 families has a radio and 187 has T.V. The number of families in that group having both a radio and \(a\) T.V is

116760 Let \(A, B, C\) be finite sets, Suppose that \(n(A)=\) \(10, \mathbf{n}(\mathrm{B})=15, \mathrm{n}(\mathrm{C})=20, \mathrm{n}(\mathrm{A} \cap \mathrm{B})=8\) and \(n(B \cap C)=9\). Then the possible value of \(\mathbf{n}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})\) is