119272 If
\(\sum_{\mathrm{k}=1}^{31}\left({ }^{31} \mathbf{C}_{\mathrm{k}}\right)\left({ }^{31} \mathbf{C}_{\mathrm{k}-1}\right)-\sum_{\mathrm{k}=1}^{30}{ }^{30} \mathrm{C}_{\mathrm{k}}\left({ }^{30} \mathbf{C}_{\mathrm{k}-1}\right)=\frac{\boldsymbol{\alpha}(60 !)}{(30 !)(31 !)},\)
where \(\alpha \in R\), then the value of \(16 \alpha\) is equal to

119273 The number of numbers, strictly between 5000 and 10000 can be formed using the digits 1,3 , \(5,7,9\) without repetition, is

119274 The number of ways of selecting two numbers a and \(b, a \in\{2,4,6, \ldots .100\}\) and \(b \in\{1,3,5, \ldots 99\}\) such that 2 is the remainder when \(a+b\) is divided by 23 is

119275 The remainder when \(7^{2022}+3^{2022}\) is divided by 5 is :

119272 If
\(\sum_{\mathrm{k}=1}^{31}\left({ }^{31} \mathbf{C}_{\mathrm{k}}\right)\left({ }^{31} \mathbf{C}_{\mathrm{k}-1}\right)-\sum_{\mathrm{k}=1}^{30}{ }^{30} \mathrm{C}_{\mathrm{k}}\left({ }^{30} \mathbf{C}_{\mathrm{k}-1}\right)=\frac{\boldsymbol{\alpha}(60 !)}{(30 !)(31 !)},\)
where \(\alpha \in R\), then the value of \(16 \alpha\) is equal to

119273 The number of numbers, strictly between 5000 and 10000 can be formed using the digits 1,3 , \(5,7,9\) without repetition, is

119274 The number of ways of selecting two numbers a and \(b, a \in\{2,4,6, \ldots .100\}\) and \(b \in\{1,3,5, \ldots 99\}\) such that 2 is the remainder when \(a+b\) is divided by 23 is

119275 The remainder when \(7^{2022}+3^{2022}\) is divided by 5 is :

119272 If
\(\sum_{\mathrm{k}=1}^{31}\left({ }^{31} \mathbf{C}_{\mathrm{k}}\right)\left({ }^{31} \mathbf{C}_{\mathrm{k}-1}\right)-\sum_{\mathrm{k}=1}^{30}{ }^{30} \mathrm{C}_{\mathrm{k}}\left({ }^{30} \mathbf{C}_{\mathrm{k}-1}\right)=\frac{\boldsymbol{\alpha}(60 !)}{(30 !)(31 !)},\)
where \(\alpha \in R\), then the value of \(16 \alpha\) is equal to

119273 The number of numbers, strictly between 5000 and 10000 can be formed using the digits 1,3 , \(5,7,9\) without repetition, is

119274 The number of ways of selecting two numbers a and \(b, a \in\{2,4,6, \ldots .100\}\) and \(b \in\{1,3,5, \ldots 99\}\) such that 2 is the remainder when \(a+b\) is divided by 23 is

119275 The remainder when \(7^{2022}+3^{2022}\) is divided by 5 is :

119272 If
\(\sum_{\mathrm{k}=1}^{31}\left({ }^{31} \mathbf{C}_{\mathrm{k}}\right)\left({ }^{31} \mathbf{C}_{\mathrm{k}-1}\right)-\sum_{\mathrm{k}=1}^{30}{ }^{30} \mathrm{C}_{\mathrm{k}}\left({ }^{30} \mathbf{C}_{\mathrm{k}-1}\right)=\frac{\boldsymbol{\alpha}(60 !)}{(30 !)(31 !)},\)
where \(\alpha \in R\), then the value of \(16 \alpha\) is equal to

119273 The number of numbers, strictly between 5000 and 10000 can be formed using the digits 1,3 , \(5,7,9\) without repetition, is

119274 The number of ways of selecting two numbers a and \(b, a \in\{2,4,6, \ldots .100\}\) and \(b \in\{1,3,5, \ldots 99\}\) such that 2 is the remainder when \(a+b\) is divided by 23 is

119275 The remainder when \(7^{2022}+3^{2022}\) is divided by 5 is :