Even 3 digits

[eaakbari]

How many 3 digit numbers are there which are even and have no repeated digits? (all numbers from 0-9)

[Brent@GMATPrepNow]

How many 3 digit numbers are there which are even and have no repeated digits? (all numbers from 0-9)

Take the task and break it into stages, beginning with the most restrictive stage

Stage 1: Select last (units) digit
The digit can be 0, 2, 4, 6, or 8, so there are 5 ways to accomplish stage 1.
Stage 2: Select tens digit
Once a digit has been selected, there are 9 digits left to choose from. So there are 9 ways to accomplish stage 2.
Stage 3: Select hundreds digit
2 digits have been used, so that are now 8 ways to accomplish stage 3.

By the Fundamental Counting Principle (FCP) we can complete all 3 stages (5)(9)(8) ways (= 360 ways)

Aside: For more information about the FCP, we have a free video on the subject: https://www.gmatprepnow.com/module/gmat-counting?id=775

Important: In the above calculation, we have allowed for the possibility that a zero is selected for the hundreds position (stage 3). Since numbers beginning with 0 are not 3-digit numbers, we need to subtract from 360 all even numbers that begin with a zero. How many such numbers are there? Let's break this task into stages (beginning with the most restrictive stage).

Stage 1: Select 0 as hundreds digit.
Can be accomplished in 1 way
Stage 2: Select last (units) digit
This can be accomplished in 4 ways (2, 4, 6 or 8)
Stage 3: Select tens digit
Can be accomplished in 8 ways

By the Fundamental Counting Principle (FCP) we can complete all 3 stages in (1)(4)(8) ways (= 32 ways)

So, total number of 3-digit even numbers that do not use any digit more than once = 360 - 32 = 328

Cheers,
Brent

[sana.noor]

My appproach
3 digits which are even and have no repeated digit thus we can write digits with even unit digits as
XY2
XY4
XY6
XY8
XY0
Now start counting numbers from 102 to 198
1st pattern----->102, 104 106 108 and 120....
second pattern---> 124, 126, 128, 130
third pattern---> 132 134 136 138 140
fourth pattern----> 142 146 148 150
fifth pattern ---> 152 154 156 158 160

you can now see that when the middle digit is even we have four such numbers and when the middle digit is odd we have five such numbers. so between 102 and 198 we can count such 3 digit numbers which are even and no repeated digits
5+4+5+4+5+4+5+4+5 = 41
now we have to count such numbers till 998..so we multiply 41 with 8= 328 total 3 digit numbers

[vishugogo]

I have done the questions in 5 stages.

Stage 1

Unit digit is 0.

So Tenth and Hundred can be selected in 9*8 = 72 ways.

Stage 2

Unit digit is 2.

So Tenth and Hundred can be selected in 8*8 = 64 ways.

Stage 3

Unit digit is 4.

So Tenth and Hundred can be selected in 8*8 = 64 ways.

Stage 4

Unit digit is 6.

So Tenth and Hundred can be selected in 8*8 = 64 ways.

Stage 5

Unit digit is 8.

So Tenth and Hundred can be selected in 8*8 = 64 ways.

72+64+64+64+64 = 328.

[Matt@VeritasPrep]

My way would be close to Vishu's way:


If your units digit is 0, your hundreds digit can be any of 9 other numbers, and your tens digit can be any of the 8 that are either zero nor equal to your hundreds digit.

Units of 0: 9 * 8 * 1 = 72 ways

If your units digit is anything else, your hundreds digit is one of the 8 nonzero digits that doesn't match your units, and your tens digit is either 0 or any of the 7 digits that doesn't match your hundreds or units digit. (For example, say your units is 2. Your hundreds could be (1, 3, 4, 5, 6, 7, 8, or 9) and your tens could be 0 or anything that isn't 2 or the hundreds digit. So 8 hundreds digits, 8 tens, and 4 units.

Units Other Even: 8 * 8 * 4 = 256

256 + 72 = 328

[vipulgoyal]

Alt approach

E E E -- 4x4x3 + (Counting from left to right)
E O E -- 4x5x4 +
O E E -- 5x5x4 +
O O E -- 5x4x5= 48+80+200 = 328