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
3 answers
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.
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.
1 answer