hi all, thanks in advance for the help
How many even integers between 4000 and 7000 have 4 different digits?
So, started by knowing "even integers" so know the last of the four has to be one of 2,4,6,8 or 0. (this might be the last part of this whole question i get right).
so...(and i apologize if my terminology is wrong, but the last time i did this stuff, i'm betting was before most of you kind folk were born :( )
so i tried multiplying the possible number of digits for each place in a four digit number. for the last number i have 5 (2,4,6,8,0) possible. the fourth from the left 8(1,2,3,4,5,6,7,8,9,0 but taking out either 5 or 6 for the first and one of the evens for the last), the third 7 (1,2,3,4,5,6,7,8,9,0 but taking out either 5 or 6 for the first and one of the evens for the last and another number of the remaining for the third digit) and the first 2 (5,6)
2*7*8*5=560
but this answer isn't what my daughter's teacher came up with. any help to tell me where i went wrong would be much appreciated.
thanks again
courtney's dad
oops, meant "counting" and permutations
I was going back and noticed that this question was unanswered.
The problem is that there are restrictions at the beginning and end of the number that conflict with each other.
first of all to be between 4000 and 7000, the number has to start with a 4, 5, or 6
So let's start with
Case 1: starts with a 4, now to be even there are only 4 left (0,2,6,8)
So we would have 1x8x7x4 = 224
Case 2: starts with a 6, (same as case1
so another 224
case 3: starts with a 5, that means there are still 5 choices at the end to make it even
so 1x8x7x5 = 280
for a total sum of 729
notice we have a "conflict" when the even number starts with a 4 or 6
Your calculation ignored that.
Reiny,
Thank you so very much for taking the time to answer this one for me.
my daughter's hw has been really tough lately. i don't stick my nose in until she asks for help, unfortunately, those times are getting to be more and more ones like this.
thank you again, this is a fabulous service you all provide.
Hi Courtney's dad! It's great that you're helping your daughter with her math problem.
Let's break down the problem step by step to find where the mistake might be. You correctly identified that the last digit of the four-digit number must be even. There are five possible even digits to choose from: 2, 4, 6, 8, and 0.
Next, let's consider the third digit from the left. It should be one of the remaining digits (1, 3, 5, 7, or 9) since it needs to be different from the last digit. However, we also need to exclude the digit used for the first digit of the number. You correctly excluded 5 and 6 from the options because they are already used as the first digit. So there are 3 possible choices for the third digit.
Moving on to the fourth digit from the left, we can choose from the remaining digits (excluding the last digit and the digit used for the first and third places). This means we have 8 options to choose from (because we have already used 2, 4, 6, 8, and the third digit).
Finally, we consider the first digit. Since we have used 5 and 6 as the third digit and the digit for the first digit, there are only 2 digits remaining (1 and 3).
Now, let's multiply our choices together:
5 (choices for the last digit) * 3 (choices for the third digit) * 8 (choices for the fourth digit) * 2 (choices for the first digit) = 240.
Therefore, the correct answer is 240, not 560. It seems like there was a mistake in your calculation.
I hope this clarifies the problem for you. Let me know if you have any more questions!