How many 3 digit odd numbers can be formed by using the digits 1, 2, 3, 4, 5, 6 when the repetition of the digits is not allowed ?

Explanation

`100^(th) 10^(th)` unit
Number of ways of unit place = 3 (by 1 or 3 or 5)
Number of ways of filling 10th place = 5
Number of ways of filling `100^(th)` place = 4
`therefore` By F.P.C, Total ways `=3xx5xx4=60`
`therefore` Required number of numbers = 60.