In an examination, there are three multiple choice questions and each question has four choices. Number of ways in which a student can fail to get all answers correct, is

Explanation

Let the M and `M^(')` represent seats of the master and mistress respectively annd let `a_(1),a_(2),a_(3), . .,a_(2n)` represent the 2n seats.

Let the guests who must not be placed next to one another be called P and Q.
Now, put P at `a_(1)` and Q at any position, other than `a_(2)`, say at remaining (2n-2) positions in (2n-2)! ways. hence, there will be altogether (2n-2)(2n-2)! arrangments of the guests, when P is at `a_(1)`.
the same number of arrangements when P is at `a_(n)` or `a_(n+1)` or `a_(2n)`. thus, for these positions `(a_(1),a_(n),a_(n+1),a_(2n))` or P, there are altogether `4(2n-2)(2n-2)!` ways.
If P is at `a_(2)`, then there are altogether `(2n-3)` positions for Q. hence, there will be altogether `(2n-3)(2n-2)!`
arrangements of the guests, when P is at `a_(2)`.
the same number of arrangements can be made when P is at any other position exception the four positions.
`a_(1),a_(n),a_(n+1),a_(2n)`.
Hence, for these (2n-4) positions of P, there will be altogether
`(2n-4)(2n-3)(2n-2)!` arrangements of the guests . . . (ii)
Hence, from Eqs. (i) and (ii), the total number of ways of arranging the guests
`=4(2n-2)(2n-2)!+(2n-4)(2n-3)(2n-2)!`
`=(4n^(2)-6n+4)(2n-2)!`.