# number of even integers with no repeated digits between 20000 and 70000

• August 11th 2012, 11:36 PM
myenemy
number of even integers with no repeated digits between 20000 and 70000
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
• August 11th 2012, 11:54 PM
a tutor
Re: number of even integers with no repeated digits between 20000 and 70000
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.
• August 12th 2012, 08:11 PM
Soroban
Re: number of even integers with no repeated digits between 20000 and 70000
Hello, myenemy!

I got a different answer . . .

Quote:

How many even integers with no repeated digits between 20,000 and 70,000?

We have a five-digit number:. $ABCDE$

The first digit $A$ can be: $\{2,3,4,5,6\}$ ... 5 choices.

The last digit $E$ can be: $\{0,2,4,6,8\}$ ... 5 choices.

There are two cases to consider.

[2] $E$ = 0 or 8 . . . 2 choices.

. . Then there are 5 choices for $A.$

. . The other 3 digits can be chosen in $_8P_3 = 336$ ways.

Hence, there are: $2\cdot5\cdot336 \,=\,3360$ numbers that end in 0 or 8.

[2] $E$ = 2, 4 or 6 . . . 3 choices.

. . Then there are only 4 choices for $A.$

. . The other 3 digits can be chosen in $_8P_3 = 336$ ways.

Hence, there are: $3\cdot4\cdot336 \,=\,4032$ numbers that end in 2, 4 or 6.

Therefore, there are: $3360 + 4032 \:=\:7392$ even numbers
. . between 20,000 and 70,000 with no repeated digits.
• August 13th 2012, 12:14 AM
a tutor
Re: number of even integers with no repeated digits between 20000 and 70000

Quote:

Originally Posted by a tutor
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.

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. :)