How many 3 - digit numbers can be formed by using the digits 1 to 9 if no digit is repeated ?

Answer

`(n(n-4)(n-5))/(6)`

Explanation

Let the object be denoted by `a_(1),a_(2),a_(3), . .,a_(n)` arranged in a circle, we have to select 3 objects so that no two of them are consecutive. For this, we first find the number of ways in which 2 or 3 objects are consecutive. Now, number of ways in which 2 or 3 objects are consecutive, is obtained as follows with `a_(1)`. the number of such triples is
`a_(1)a_(2)a_(3),a_(1)a_(2)a_(4),a_(1)a_(2)a_(5), . . .,a_(1)a_(2)a_(n-1)`.
[Since, we have excluded `a_(1)a_(2)a_(n),` so it will be repeated again. if we start with `a_(n)`, then we shall get triples: `a_(n)a_(1)a_(2),a_(n)a_(1)a_(3)`]
so, number of such triples when we start with `a_(1)`, is (n-3). similarly, with `a_(2),a_(3),a_(4), k, . . .,` we shall get the numbers of triples that is (n-3).
but total number of triples is `.^(n)C_(3)`.
Hence, required number of ways `=.^(n)C_(3)-n(n-3)`
`=(n(n-1)(n-2))/(1*2*3)-n(n-3)=(n)/(6)[n^(2)-3n+2-6n+18]`
`=(n)/(6)(n^(2)-9n+20)=(n)/(6)(n-4)(n-5)`.