Find out how many distinct three-digit numbers can be formed using all the digits of 1, 2, and 3.

Explanation

Given, total different things=n
the number of permutations of n things taken one at a time `=.^(n)P_(1)=hn`, now if we taken two at a time (repetition is allowed), then first place can be filled by n ways and second place can again be filled in n ways.
`therefore`The number of permutations of n things takenn two at a time
`=.^(n)P_(1)xx.^(n)P_(1)=nxxn=n^(2)`
Similarly, the number of permutations of n things taken three at a time`=n^(3)`
The number of permutations of n things taken r at a
time `=n^(r)`. hence, the total number of permutations
`=n+n^(2)+n^(3)+ . . .+n^(r)`
`=(n(n^(r)-1))/((n-1))` [sum of r terms of a GP]