+0  
 
0
29
5
avatar

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
Sort: 

5+0 Answers

 #1
avatar+74649 
+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

 

 

cool cool cool

CPhill  Jun 14, 2017
 #2
avatar
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
avatar
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
 #4
avatar+74649 
0

 

Sorry about that....I mis-read the question.....

 

 

cool cool cool

CPhill  Jun 14, 2017
 #5
avatar+74649 
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!!! "

 

 

cool cool cool

CPhill  Jun 14, 2017

12 Online Users

avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details