Distinct-digit Even Numbers

Call the three digit number is $N=\overline{abc}$ . $c$ is one of digits $0,2,4,6,8$ . $b$ is one of digits $0,1,2,3,4,5,6,7,8,9$ . a is one of the digits $1,2,3,4,5,6,7,8,9$ . Since $a,b,c$ are distinct from each other,

Case 1: $c=0$ . There are $9\times 8$ ways.

Case 2: $c=2, 4, 6, 8$ . $8 \times 8$ ways.

In total, there are $8\times 8\times 4+9\times 8 =328$ numbers.

[Latex edits - Calvin]

even numbers only have five digits in the end (i.e 0,2,4,6,8) , so there are five conditions : 1: when last number is zero ( 0 ), and starting with 1, there are 8 possible 3-digit numbers and so on till the numbers starting with 9 there are 8 possible 3-digit numbers as we are skipping the number with same middle digit as first one therefore there are 8 possible numbers so , 9 x 8 = 72 2: when last no. is 2, and starting with 1, there are 8 possible 3-digit numbers and so on till the numbers starting with 9 there are 8 possible 3-digit numbers but this time we will skip the series of no.s starting with 2 as the digits should not repeat so, 8 x 8 =64 3: when last no. is 4, and starting with 1, there are 8 possible 3-digit numbers and so on till the numbers starting with 9 there are 8 possible 3-digit numbers but this time we will skip the series of no.s starting with 4 as the digits should not repeat so, 8 x 8 =64 4: when last no. is 6, and starting with 1, there are 8 possible 3-digit numbers and so on till the numbers starting with 9 there are 8 possible 3-digit numbers but this time we will skip the series of no.s starting with 6 as the digits should not repeat so, 8 x 8 =64 5: when last no. is 8, and starting with 1, there are 8 possible 3-digit numbers and so on till the numbers starting with 9 there are 8 possible 3-digit numbers but this time we will skip the series of no.s starting with 8 as the digits should not repeat so, 8 x 8 =64 so the total becomes , 72 + 64 + 64 + 64 + 64 = 328 So 328 is the ANSWER

There are 5 digits for unit place i.e. 0, 2, 4, 6, 8. Work the solution for each UNIT place separately.

Lets take 0 for UNIT place first. After using 0; we are left with 9 digits to work with for the HUNDRED place. after putting any of the digit in the hundred place, we are left with 8 digits for TEN place to work with. so the situation is like this:

9 x 8 x 1 = 72

Now we take 2 for UNIT place. which means we are again left with 9 digits for HUNDRED place. but since we can't put 0 here, so we are left with 8 digits. after putting one digit in hundred place we are left with 7 digits and a 0 for TEN place. which means we have 8 choices for TEN place. so the situation is:

8 x 8 x 1 = 64

For the digits 4, 6, and 8; we will work similarly as we worked with 2 in UNIT place. And at the end adding all the individual answers give us:

72 + 64 + 64 + 64 + 64 = 328

which is our answer!

in this problem,we've to know there are total 900 three digit numbers. 100_ 199 200_ 299 300_ 399 400_ 499 500_ 599 600_ 699 700_ 799 800_ 899 900_ 999 in other hand, we've to check out, which one is start with even digit or odd digit. like (100 199), here 100 start with odd digit i.e.1 & (200 299) here 200 is start with even digit i.e. 2. now we can calculate in (100 199), total number of positive even integers are 40, but in (200 299), total number of positive even integers are 32.(we can also calculate it by forming a pattern).here there are total 5 column, which are starting with odd digit 100_ 199 300_ 399 500_ 599 700_ 799 900_ 999 so, we'll multiply 40 with 5 (40 /times 5=200) but there are 4 column starting with even integers, which are... 200_ 299 400_ 499 600_ 699 800_ 899 so, we'll multiply 32 with 4 (32 /times 4=128) so, by adding 200+128=328(answer)

We see that since the number has to be even, the last digit can be 2,4,6,8,0. we take 2 cases: case 1: last digit is 0. now the options available for the first digit=9(all digits, except 0.) and options for the 2nd digit is 8. (0 and the 1st digit cannot repeat.) so case 1 has 9 8=72 numbers. Case 2: last digit =2,4, 6 or 8. so 1st digit cannot be 0(then the number wouldn't be 3 digit anymore) and the last digit cannot repeat. so options for the 1st digit=8. and 2nd digit cannot have the 1st digit or the 3rd digit. but can have 0. so options=8. and there are thus 8^2=64 numbers for each of 2,4,6,8 in the unit's place. so case 3 has 64 4=256 numbers. now by addition principle case 1 +case 2=328(answer)

Call one of our numbers \overline{abc}

If (a is odd (5 choices), b is odd (4 choices) and c is even (5 choices), we have 5 \times 4 \times 5 = 100 numbers

Else if (a is odd (5 choices), b is even (5 choices) and c is even (4 choices), we have 5 \times 5 \times 4 = 100 numbers

Else if (a is even (4 choices), b is even (4 choices) and c is even (3 choices), we have 4 \times 4 \times 3 = 48 numbers

Else if (a is even (4 choices), b is odd (5 choices) and c is even (4 choices), we have 4 \times 5 \times 4 = 80 numbers

We have a total of (100+100+48+80) = 328 numbers

http://www.brilliantscholars.com/assessment/s/geometry-and-combinatorics/86576/#

This event can be partitioned into two mutually exclusive cases.

Case 1. The number ends in 0. In this case, there are nine choices for the first digit (1—9) and then eight for the second (since 0 and the first digit must be excluded). So there are 9 * 8 = 72 numbers of this type.

Case 2. The number does not end in 0. Now there are four choices for the final digit (2, 4, 6 and 8), then eight choices for the first digit (0 and the last digit are excluded), and eight choices for the second digit (the first and last digits are excluded). There are 4 * 8 * 8 = 256 numbers of this type.

By the addition rule, there are 72 + 256 = 328 even numbers in the range 100—999 with no repeated digits.

fix unit digit as even like 0 2 4 6 8 now 1)- - 0= 9 8 2)- - 2= 8 8 3)- - 4= 8 8 4)- - 6= 8 8 5)- - 8= 8 8 now add these ways which comes to 328 9 8 + 8 8 + 8 8 + 8 8 + 8 8 =328

Case 1: Last digit is not 0. There are 4 possibilities for the last digit (2, 4, 6, 8), 8 possibilities for the first digit (not the last, and not 0), and then 8 possibilities for the second digit (not the last, not the first). Hence, by the rule of product, there are $4 \times 8 \times 8 = 256$ such 3 digit even integers.

Case 2: Last digit is 0. There are 9 possiblities for the first digit (only can’t be 0), and 8 possibilities for the second digit. Hence, by the rule of product, there are $9 \times 8 = 72$ possibilities.

These cases are clearly distinct, since the last digit is different. Thus, by the rule of sum, there are $256 + 72 = 328$ 3 digit even integers, with all distinct digits.