How many 5-digit even numbers can be made using the digits 2, 4, 6, 7, and 8, if digits cannot be repeated?
none. You only show 4 even digits.
there is 5 digits, you count wrong.
Vincent -- you don't read very carefully. Steve said that you only show 4 even digits.
Actually, I was in error in my answer. Only the last digit has to be even!!
So, there are
4*4*3*2*1 = 96 choices.
Sorry, my bad.
To find the number of 5-digit even numbers that can be made using the digits 2, 4, 6, 7, and 8 without repeating digits, we can break the problem into steps:
Step 1: Count the total number of digits available.
In this case, we have 5 digits available: 2, 4, 6, 7, and 8.
Step 2: Determine the options for the most significant digit (the leftmost digit).
Since the number needs to be even, the most significant digit cannot be 7. So we have 4 options: 2, 4, 6, and 8.
Step 3: Determine the options for the remaining four digits (after the most significant digit is chosen).
For the second digit, any of the remaining 4 digits can be chosen (since repetition is not allowed).
For the third digit, any of the remaining 3 digits can be chosen.
For the fourth digit, any of the remaining 2 digits can be chosen.
For the fifth (and final) digit, only 1 digit is left.
Step 4: Multiply the number of options for each step together.
To get the total number of 5-digit even numbers, we multiply the number of options for each step together:
4 (options for most significant digit) × 4 (options for second digit) × 3 (options for third digit) × 2 (options for fourth digit) × 1 (option for fifth digit) = 96
Therefore, there are 96 different 5-digit even numbers that can be made using the digits 2, 4, 6, 7, and 8, without repeating any digit.