First digit even 3 ways, last digit even 4 ways, other digits 8*7*6 ways.
First digit odd 3 ways, last digit even 5 ways, other digits 8*7*6 ways.
I have been given the problem to solve how many even integers with no repeated digits between 20,000 and 70,000.
my first thought was to have the set X = {0,1,2,3,4,5,6,7,8,9}
and then propose there are 5 positions in which these numbers can be placed, being ABCDE
therefore E has to be either {0,2,4,6,8} => |E| = 5
D is any other of the 10 ten numbers excluding E => |D| = 8 and so on.
therefore i originally thought that the answer would be 5x9x8x7x2 (using the mulitplication prinicple)
As A is either 2,3,4,5,6,7 and cant be what B,C,D was therefore |A| = 2, i know that this is not true as BCD can be numbers outside of this region. but i am unclear as to how to find |A| without knowing what values B,C,D took on? any help would be appreciated.
Thanks
Reece
Hello, myenemy!
I got a different answer . . .
How many even integers with no repeated digits between 20,000 and 70,000?
We have a five-digit number:.
The first digit can be: ... 5 choices.
The last digit can be: ... 5 choices.
There are two cases to consider.
[2] = 0 or 8 . . . 2 choices.
. . Then there are 5 choices for
. . The other 3 digits can be chosen in ways.
Hence, there are: numbers that end in 0 or 8.
[2] = 2, 4 or 6 . . . 3 choices.
. . Then there are only 4 choices for
. . The other 3 digits can be chosen in ways.
Hence, there are: numbers that end in 2, 4 or 6.
Therefore, there are: even numbers
. . between 20,000 and 70,000 with no repeated digits.
Thanks Soroban. I made a silly little mistake. It can't start with a 7.
First digit even 3 ways, last digit even 4 ways, other digits 8*7*6 ways.
First digit odd 2 ways, last digit even 5 ways, other digits 8*7*6 ways.
So 3.4.8.7.6 + 2.5.8.7.6 = 7392 in agreement with Soroban.