Let's solve the problems step by step:
### Part i) Finding the Total Number of Different 5-Digit Numbers
To form a 5-digit number using five out of the eight digits 1, 2, 3, 4, 5, 6, 7, 8 without repeating any digit, we need to determine how many such arrangements are possible.
This is a problem of permutations of a subset, where we have to choose and arrange [tex]\( k = 5 \)[/tex] digits out of [tex]\( n = 8 \)[/tex] digits.
The formula for permutations of choosing [tex]\( k \)[/tex] items from [tex]\( n \)[/tex] items is:
[tex]\[ P(n, k) = \frac{n!}{(n-k)!} \][/tex]
Apply this formula:
[tex]\[ P(8, 5) = \frac{8!}{(8-5)!} = \frac{8!}{3!} \][/tex]
Now, calculating the factorial values (without intermediate steps per instruction):
[tex]\[ 8! = 40320 \][/tex]
[tex]\[ 3! = 6 \][/tex]
Thus:
[tex]\[ P(8, 5) = \frac{40320}{6} = 6720 \][/tex]
So, the total number of different 5-digit numbers that can be formed using the digits 1 to 8 is 6720.
### Part ii) Finding the Number of 5-Digit Numbers Greater Than 60,000
To find how many of these 5-digit numbers are greater than 60,000, we need to consider the positional significance of the digits. Specifically, for a number to be greater than 60,000, its first digit must be either 6, 7, or 8.
#### Case when the first digit is 6, 7, or 8:
1. We have 3 choices (6, 7, or 8) for the first digit.
2. After choosing one digit as the first, we are left with 7 digits and need to arrange the remaining 4 digits.
Hence, the number of ways to arrange these 4 digits from the remaining 7 is given by:
[tex]\[ P(7, 4) = \frac{7!}{(7-4)!} = \frac{7!}{3!} \][/tex]
Using the previously calculated factorials:
[tex]\[ 7! = 5040 \][/tex]
[tex]\[ 3! = 6 \][/tex]
Thus:
[tex]\[ P(7, 4) = \frac{5040}{6} = 840 \][/tex]
Since the first digit can be any of 6, 7, or 8:
[tex]\[ 3 \times 840 = 2520 \][/tex]
So, the number of 5-digit numbers greater than 60,000 is 2520.
### Summary
1. Total different 5-digit numbers: 6720
2. 5-digit numbers greater than 60,000: 2520
