How many 7 digit even numbers can be formed using all of the digits 0,1,1,2,3,4?

The answer is 1320. I tried 7!/2! but I know its supposed to be even, so it's not right to do 2!. How do I get to 1320?

let's first of all assume that the two 1's are distinguishable, e.g. one is red, the other blue

case1, our even number ends in a zero
number of such cases = 6x5x4x3x2x1x= 6! = 720
but the two 1's ARE distinguishable, so
the number of evens that end in a zero are 720/2 = 360

case2, our even number ends in either 2 or 4
and we of course can't start with a zero.
fill the end first
number of ways for our number
= 2 x 5 x 5 x 4 x 3 x 2 x 1 = 1200
taking care of the same 1's
number of case2 numbers = 1200/2 = 600

total number = 600 + 360 = 960

I disagree with the answer of 1320

To determine the number of 7-digit even numbers that can be formed using the digits 0, 1, 1, 2, 3, and 4, we need to break down the problem into smaller steps.

Step 1: Find the total number of possible 7-digit numbers ignoring the restriction of being even.

In this case, we have a total of 6 digits to choose from (0, 1, 1, 2, 3, and 4). Since repetition is allowed, we have 6 choices for each digit. Thus, the total number of possible 7-digit numbers is 6 multiplied by 6 multiplied by 6 multiplied by 6 multiplied by 6 multiplied by 6 multiplied by 6, which can be represented as 6^7.

Step 2: Adjust the total count to account for the restriction of the numbers being even.

An even number always ends in 0, 2, or 4.
- If the last digit is 0, then we have no restrictions for the remaining 6 digits.
- If the last digit is 2, then we can choose any of the remaining 5 digits for the first digit (it cannot be 0), and similarly, we have no restrictions for the remaining 5 digits.
- If the last digit is 4, the same logic applies, but we have 4 choices for the second digit (1, 1, 3, and 4) since 2 has already been used.

Therefore, the total number of 7-digit even numbers is equal to:
6^7 - 5^7 - 4 * 5^6

Calculating this expression, we get:
279,936 - 78,125 - 4 * 15,625 = 279,936 - 78,125 - 62,500 = 139,311.

Hence, the correct answer is 139,311, not 1320.

To find the number of 7-digit even numbers that can be formed using the digits 0, 1, 1, 2, 3, and 4, we need to break the problem down into steps.

Step 1: Count the total number of ways to arrange all 7 digits.
Since we have 7 distinct digits, we can arrange them in 7! ways (7 factorial). However, in this case, we have two identical digits "1". So, we need to divide by 2! to eliminate the overcounting from the repeated digit.
Total arrangements = 7! / 2!

Step 2: Determine the restriction for the last digit.
In order for the number to be even, the last digit must be one of 0, 2, or 4. There are no restrictions on the remaining 6 digits.

Step 3: Calculate the number of even numbers.
To find the total number of even numbers, we need to consider each of the three possibilities for the last digit (0, 2, or 4) and multiply it by the number of allowable arrangements for the remaining 6 digits.

Case 1: Last digit = 0
Since the last digit is fixed, we have 6 digits remaining to arrange. So, the number of valid arrangements is (6! / 2!).

Case 2: Last digit = 2
Similarly, we have 6 digits remaining to arrange, giving us (6! / 2!) valid arrangements.

Case 3: Last digit = 4
Once again, we have 6 digits left to arrange, resulting in (6! / 2!) valid arrangements.

Step 4: Calculate the total number of even numbers.
Add the results from the three cases to get the total number of even numbers:
Total = (6! / 2!) + (6! / 2!) + (6! / 2!)

Now we can simplify the expression:
Total = 3 * (6! / 2!)

Using the formula for factorials, we have:
Total = 3 * (6 * 5 * 4 * 3 * 2 * 1) / (2 * 1)

Cancelling out common factors:
Total = 3 * (6 * 5 * 4 * 3)
Total = 3 * 360
Total = 1080

Therefore, there are 1080 seven-digit even numbers that can be formed using the digits 0, 1, 1, 2, 3, and 4.

Sorry, the digits are 0,1,1,2,3,4,5