InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Prove that `((n^(2))!)/((n!)^(n))` is a natural number for all n `in` N. |
|
Answer» We have to prove that `(n^(2))!` is divisible by `(n!)^(n)`. We know that product of r consecutive integers is divisible by r! Now `(n^(2))!=1xx2xx3xx4xx..xx n^(2)` `=(1xx2xx3xx..xx n)xx` `[(n+1)(n+2)xx.. .. Xx(2n)]xx` `[(2n+1)(2n+2).. .. (3n)]xx` .. .. .. .. `[(n^(2)-(n-1)(n^(2)-n).. .. n^(2)]` Thus `(n^(2))!` consists n groups of product of n consecutive integers. Each group is divisible by n!. So, `(n^(2))!` is divisible by `(n!)^(n)`. |
|