Let `f (n) = |{:(n,,n+1,,n+1),(.^(n)P_(n),,.^(n+1)P_(n+1),,.^(n+2)P_(n+2)),(.^(n)C_(n),,.^(n+1)C_(n+1),,.^(n+2)C_(n+2)):}|` where the sysmbols have their usual neanings .then f(n) is divisible by

Options

A. `n^(2)+n+1`
B. `(n+1)!`
C. `n!`
D. none of these

Answer

A, C

Explanation

`f(n) = |{:(n,,n+1,,n+2),(n!,,(n+1)!,,(n+2)!),(1,,1,,1):}|=|{:(n,,1,,1),(n!,,nn!,,(n+1)(n+1)!),(1,,0,,0):}|`
[Applying `C_(3) to C_(3) -C_(2) " and " C_(2) to C_(2)-C_(1)]`
`=(n+1)(n+1)!-nn! =n![(n+1)^(2)-n]`
`=n!(n^(2) +n+1)` ltbr. Thus f(n) is divisible by n! and `n^(2) +n+1`