3. Using permutation or otherwise prove that $\frac { n ^ { 2 } ! } { ( n ! ) ^ { n } }$ is an integer, where $n$ is a positive integer. Sol. Let there be $\mathrm { n } ^ { 2 }$ objects distributed in n groups, each group containing n identical objects. So number of arrangement of these $n ^ { 2 }$ objects are $\frac { n ^ { 2 } ! } { ( n ! ) ^ { n } }$ and number of arrangements has to be an integer. Hence $\frac { \mathrm { n } ^ { 2 } } { ( \mathrm { n } ! ) ^ { \mathrm { n } } }$ is an integer.
If $y$ is a function of $x$ and $\ln ( x + y ) - 2 x y = 0$, then the value of $y ^ { \prime } ( 0 )$ is equal to
3. Using permutation or otherwise prove that $\frac { n ^ { 2 } ! } { ( n ! ) ^ { n } }$ is an integer, where $n$ is a positive integer.
Sol. Let there be $\mathrm { n } ^ { 2 }$ objects distributed in n groups, each group containing n identical objects. So number of arrangement of these $n ^ { 2 }$ objects are $\frac { n ^ { 2 } ! } { ( n ! ) ^ { n } }$ and number of arrangements has to be an integer.\\
Hence $\frac { \mathrm { n } ^ { 2 } } { ( \mathrm { n } ! ) ^ { \mathrm { n } } }$ is an integer.\\