How many 5 digits numbers less than 60000 and even can be formed using the digits 1,2,4,5,7 and 9
This is tricky since we have over-lapping restrictions, so let's use cases
We can only use 1,2,4 or 5 at the front, (it has to be < 60000)
case1: it starts with 1 or 5
number of ways = 2*4*3*2*2 , the last 2 is for the 2 evens we can use
= 96
case2: it starts with a 2,
number of ways = 1*4*3*2*1 , the last can only be the 4
= 24
case3: it starts with a 4, which is the same as case2
number of ways = 96+24+24 = 144
Oh, let me use my number-crunching clown calculator for this one. Give me a moment to gather all the digits... 1, 2, 4, 5, 7, and 9. Okay, let's start assembling some numbers!
To form a 5-digit number less than 60000, we have a few conditions to meet:
1. The first digit cannot be 7 or 9 because that would exceed 60000.
2. The first digit cannot be 1 or 2 because we want an even number.
So, let's go digit by digit, clown-style!
For the first digit, we have only two options: 4 or 5.
For the second digit, we have five remaining options: 1, 2, 4, 5, or 7.
For the third, fourth, and fifth digits, we have six remaining options each (since we can reuse digits).
Multiplying these together, we get:
2 (for first digit) * 5 (for second digit) * 6 (for third digit) * 6 (for fourth digit) * 6 (for fifth digit) = 2 * 5 * 6 * 6 * 6 = 720.
So, using these digits, we can form 720 5-digit numbers less than 60000 that are even. There you go, my number-crunching clown calculations at your service!
To determine the number of 5-digit even numbers that can be formed using the digits 1, 2, 4, 5, 7, and 9, we need to consider the following:
1. The first digit cannot be zero, as we want 5-digit numbers.
2. The last digit must be even (either 2 or 4) because an even number must end with an even digit.
Now, let's solve this step-by-step:
Step 1: Determine the possible options for the first digit.
Since the first digit cannot be zero, we have 5 options (1, 2, 4, 5, and 7).
Step 2: Determine the possible options for the middle three digits.
We have 5 digits available for each of the three middle positions. So, for each position, we have 5 options. Therefore, for the three middle positions combined, we have 5 × 5 × 5 = 125 options.
Step 3: Determine the possible options for the last digit.
The last digit must be 2 or 4, so we have 2 options.
Step 4: Multiply the results from steps 1, 2, and 3 to get the total number of possibilities.
The total number of possibilities is 5 (options for the first digit) × 125 (options for the middle three digits) × 2 (options for the last digit) = 1250.
Therefore, there are 1250 five-digit even numbers that can be formed using the digits 1, 2, 4, 5, 7, and 9, and are less than 60000.
To find the number of 5-digit numbers less than 60000 and even that can be formed using the digits 1, 2, 4, 5, 7, and 9, we need to break down the problem into smaller steps.
Step 1: Determine the first digit
Since the number needs to be less than 60000, the first digit cannot be 6. Therefore, the first digit can only be either 1, 2, 4, 5, or 7.
Step 2: Determine the second digit
Since the number is a 5-digit number, and it needs to be even, the second digit can only be 2, 4, or 6. However, we already established that the first digit cannot be 6, so the second digit can only be 2 or 4.
Step 3: Determine the remaining digits
For the remaining three digits, any of the digits 1, 2, 4, 5, 7, and 9 can be used since there are no restrictions on their values.
Step 4: Calculate the total number of possibilities
Multiply the number of options for each step together to find the total number of possibilities:
- Number of options for the first digit: 5 (as there are 5 choices: 1, 2, 4, 5, or 7)
- Number of options for the second digit: 2 (as there are 2 choices: 2 or 4)
- Number of options for the remaining three digits: 6 × 6 × 6 = 216 (as there are 6 choices for each digit)
Multiply these numbers together: 5 × 2 × 216 = 2160
Therefore, there are 2160 five-digit numbers less than 60000 and even that can be formed using the digits 1, 2, 4, 5, 7, and 9.