How many 4 digit codes can we make provided that
-the codes contain of numbers only
-codes may not equal to 4000
-codes may not be less than 4000
-numbers may be repeated
-codes have to be even numbers??????
To find the number of 4-digit codes that satisfy these conditions, we can break down the problem step by step:
Step 1: Count the number of valid choices for each digit position.
- For the first digit position, since codes may not be less than 4000, we have the following choices: 4, 5, 6, 7, 8, 9 (6 choices).
- For the second, third, and fourth digit positions, since numbers may be repeated and codes have to be even, we have the following choices: 0, 2, 4, 6, 8 (5 choices).
Step 2: Multiply the number of choices for each digit position.
Since each digit position is independent of the others, we can multiply the number of choices from step 1 together to get the total number of valid codes:
Total number of valid codes = number of choices for first digit position * number of choices for second digit position * number of choices for third digit position * number of choices for fourth digit position.
Total number of valid codes = 6 * 5 * 5 * 5 = 750.
So, there are 750 valid 4-digit codes that satisfy the given conditions.