How many three-digit even numbers can be formed using the digits from 0-9 if each digit can only be used once?

Guest Jun 14, 2017

#1**+1 **

We have 10 ways to choose the first number, 9 ways to choose the second and 8 ways to choose the third

So

10 * 9 * 8 = 720 possible combinations

CPhill Jun 14, 2017

#2**0 **

err.... sorry to disagree,but don't we start with

10

C

3 to begin and then decide how many of these are even? I think this one's a bit more complicated than first meets the eye -----

Guest Jun 14, 2017

#3**0 **

How many three-digit** even numbers.......etc.**

Since a 3-digit number cannot begin with zero, then we have 9 numbers to choose from. And we have 10 digits to choose from for the 2nd number. And since he/she wants an EVEN 3-digit numbers, which means they must end in: 0, 2, 4, 6, 8 or five numbers, therefore we should have:

9 x 10 x 5 =450 numbers.

Guest Jun 14, 2017

#5**0 **

However....I'm sure that the answer isn't 450, either....

Assuming that we can't begin with 0, there are 900 integers from 100 - 999 and half of these will be even = 450....and some of those will have repeated digits.....so.....the answer is less than 450

Here's a "brute force" method

Let's look at all the integers that end in 0

There are 10 of these in each group of 100.....and we have 9 groups of 100 from 100-999

So....that's 10 * 9 = 90 possible integers.....but...we have to discard 10 of these (100, 200, 300. etc.) because of the repeated 0....so......that makes 80 possible numbers ending in 0

Now.....let's look at all the numbers ending in 2

Again there are 10 of these in each group of 100.....so, again, that's 90 possible numbers

But.....from these we must subtract all the numbers that have 2 as a middle digit and 2 as an ending digit.....and there are 9 of these (122, 222, 322, 422, etc )...so that gives us 81 possible numbers....

Also.....we have to subtract the numbers that start with 2 and have 2 as an ending digit....there are 9 of these, as well (202, 212, 232, 242, etc. ).....(note that we have already counted '222")....so.....the possible numbers ending in 2 with no repeated digits = 90 - 9 - 9 = 72

And the other numbers ending in 4,6 or 8 will similarly occur 72 tmes each

So.....the total combinations ending in an even number with no repeated digits = 80 + 4 * 72 =

80 + 288 = 368

As Melody always says, "That's what I think!!! "

CPhill Jun 14, 2017