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Here's why: If we first choose zero for the units digit, then the hundreds digit can be 1,2,3,4,5,6,7,8 or 9 (9 possible values)
But, if we DON'T first choose zero for the units digit, then the hundreds digit cannot be zero, since the resulting number wouldn't be a 3-digit number (e.g., 046) is not a 3-digit number). Also, the hundreds digit cannot be the same as the units digit (since all 3 digits are DIFFERENT). So, if we DON'T first choose zero for the units digit, then the hundreds digit can have only 8 possible values.
Given this, we must treat each case separately.
Case A: We choose zero for the units digit Take the task of creating 3-digit integers, and break it into stages.
We’ll begin with the most restrictive stage.
Stage 1: Select a units digit Since the units digit MUST be 0 for this case, we can complete stage 1 in 1 way
Stage 2: Select a hundreds digit The hundreds digit can be 1,2,3,4,5,6,7,8 or 9 So, we can complete stage 2 in 9 ways
Stage 3: Select a tens digit The tens digit can be any of the 10 digits EXCEPT for 0 and the number selected for the hundreds digit So, we can complete stage 3 in 8 ways
By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus create a 3-digit number) in (1)(9)(8) ways (= 72 ways) ---------------------------------- Case B: We DO NOT choose zero for the units digit
Stage 1: Select a units digit Since the units digit must be EVEN (but can't be 0), the units digit can be (2,4,6 or 8) So, we can complete stage 1 in 4 ways
Stage 2: Select a hundreds digit There are 10 digits in total. HOWEVER, the hundreds digit cannot be zero, AND the hundreds digit cannot be the same as the units digit (since all 3 digits are DIFFERENT) So, we can complete stage 2 in 8 ways
Stage 3: Select a tens digit The tens digit can be any of the 10 digits EXCEPT for the two digits selected for the hundreds and units place So, we can complete stage 3 in 8 ways
By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus create a 3-digit number) in (4)(8)(8) ways (= 256 ways) ----------------------------------------
So, the TOTAL number of possible 3-digit numbers = 72 + 256 = 328
Re: How many different 3-digit even numbers can be formed
[#permalink]
17 Sep 2022, 06:52
2
Another way to solve it this: Count the number of ways a zero can be the hundred's digit. Subtract this number from the total number of ways.
First, total number of ways: Put an even digit in the unit's place: this can be done in 5 ways (0, 2, 4, 6, 8). Put any remaining digit in the ten's place: this can be done in 9 ways. Put any remaining digit in the hundred's place: this can be done in 8 ways.
Total: 5x9x8 = 360
Second, count the number of ways a zero can be the hundred's digit:
Put an even digit in the unit's place: this can be done in 4 ways (2, 4, 6, 8). Remember, we will put 0 in the hundred's place. Put any remaining digit in the ten's place: this can be done in 8 ways. Put any a zero in the hundred's place: this can be done in 1 ways.