i)

Using a simpler example to illustrate :

When n = 4, the number will be _ _ _ _

From 0 to 9, there are 10 digits to choose from for each number place.

Number of ways for first digit = 9

(Only 1 to 9 is possible. We cannot have 0 beause that would mean it isn't a 4-digit number anymore eg. 0123 would just be 123)

Number of ways for second digit = 9
(Cannot include the same digit as the first)

Number of ways for third digit = 9
(Cannot include same digit as the second)

Number of ways for fourth digit = 9
(Cannot include same digit as the third)


Total Number of ways = 9×9×9×9 = 9⁴

So for i),

Number of ways that fit the condition = 9ⁿ

Required probability

= Number of ways that fit the condition ÷ total number of ways

= 9ⁿ ÷ 10ⁿ (there are 10 ways for each number place, so for n digits it is 10 × 10 × 10 × 10... n times)

= (9/10)ⁿ

ii)


Number of ways for first digit = 9
(Cannot include 0)


Number of ways to choose 3 number places out of the remaining n-1 places to be '0' = (n-1) C 3

= (n - 1)! / (n - 1 - 3)!3!

= (n - 1)! / (n - 4)!3!

= (n - 1)(n - 2)(n - 3) / 6

Number of ways for the remaining n - 4 number places

= 9ⁿ-⁴ (9 ways for each place since we cannot use 0)


Total number of ways possible

= 9(9ⁿ-⁴)(n - 1)(n - 2)(n - 3) / 6
= 9ⁿ-³(n - 1)(n - 2)(n - 3) / 6
= 9ⁿ (n - 1)(n - 2)(n - 3) / (6 × 9³)
= 9ⁿ (n - 1)(n - 2)(n - 3) / 4374



Required probability

= 9ⁿ (n - 1)(n - 2)(n - 3) / 4374 ÷ 10ⁿ

= (9/10)ⁿ (n - 1)(n - 2)(n - 3) / 4374