Take the task and break it into stages, beginning with the most restrictive stageeaakbari wrote:How many 3 digit numbers are there which are even and have no repeated digits? (all numbers from 0-9)
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
