How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that
repetition of the digits is not allowed ?

Options

A. `3|(N-1)`
B. `n|(N_1)`
C. `(n+1(|(N-1)`
D. `3n(n+1)|(N-1)`

Answer

A, B, C, D

Explanation

`becausex+y+z=3n`
`impliesN=`Coefficient of `alpha^(3n)` in `(1+alpha+alpha^(2)+ . . .+alpha^(2n))^(3)`
=Coefficient of `alpha^(3n)` in `(1-alpha^(2n+1))^(3)(1-alpha)^(-3)`
=Coeffieint of `alpha^(3n)` in `(1-3alpha^(2n-1))(1+.^(3)C_(1)alpha+ . . .)`
`=.^(3n+2)C_(3n-3)*.^(n+1)C_(n-1)`
`=.^(3n+2)C_(2)-3*.^(n+1)C_(2)=3n^(2)+3n+1`
`thereforeN-1=3n(n+1)`