So, 1"s place-1 10"s place-9 100"s place-7 (2 digits are already consumed and 0 can"t be used)

So 7*9*1.Im i doing the right thing?


First find the even numbers that are ending up in zeroso, for no. ending up with zero arezero at one"s place so 1 combinationnow 9 at hundreds and 8 at tensNow even numbers not ending with zero i.e. ending in 2,4,6,8so 4 at one"s place only one is used so 8 at hundreds place since 0 can not be used and one number is 0 and now 8 are left for the tens placeso, total no. of digit sequences would be=>9*8*1+8*8*4=72+256=328When zero is used up there is no problem for using up tens or hundreds but when zero is not used up after putting one of 2,4,6,8 numbers at one"s place we have to check if 0 is ending up in the most significant place or not. To avoid that we start with hundredth place. So we ensure this first.

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Here we consider numbers of the form xyz, where each of x, y, z represents a digit under the given restrictions. Since xyz has to be even, z has to be 0, 2, 3, 4, 6, or 8.

If z is 0, then x has 9 choices.

If z is 2, 4, 6 or 8 (4 choices) then x has 8 choices. (Note that x cannot be zero)

Therefore, z and x can be chosen in (1 × 9) + (4 × 8) = 41 ways. For each of these ways, y can be chosen in 8 ways.

Hence, the desired number is 41 × 8 = 328 numbers 3-digit even numbers exist with no repetitions.


Units digit can be among 0,2,4,6,8.

CASE A: If 0 is at units place, No. of terms possible is 1x9x8 = 72

CASE B: If 0 is not at units place but at hundred"s place No of terms possible is 4x1(0)x8 = 32CASE C: If 0 is not there at all, No of terms possible is: 4x8x7 = 224

total no"s = 224+32+72 = 328


eginalign & underline extcase 1 ext : All possible even 3-digit numbers are given by = $5*9*8 =360$ \ & underline extcase 2 ext : All even number with ""0"" at hundreds ext place ext are given by = $4*8*1=32$ \ & underlineThe ext number of even 3 digit is = case1 -case 2 = 360-32=328 endalign

Total three digits numers are 900

We have total 10 digits which are 0,1,2,3,4,5,6,7,8,9

But a leading zero does not have any value so we cant have zero at 100th place

So for 100th place we have 9 digits, for tenth place we have 10 digits and for units place we have 10 digits

So total 3 digits number are 9*10*10=900

If we talk about even numbers then we have only 5 digits fot units place 0,2,4,6,8So total 3 digits even numbers would be 9*10*5=450


There are some good answers here. But I just want to mention one more approach:

First calculate total 3 digit numbers without repetition: 9 x 9 x 8 = 648Then calculate 3 digit odd numbers without repetition: 8 x 8 x 5 = 320Now subtract the number of odd numbers from total: 648 - 320 = 328
On the number line the 3 digit numbers are : 100 - 999So if I would start at 1 - 999 then I have 999 numbers in total.from those I will take away the one digit numbers : 9and the two digit numbers: 90I.e: 999-99 = 900 3 digit numbers.

Now to count the even numbers: we start at 100 and skip count by 2.Or perhaps we can divide by 2: 900/2 = 450 even numbers.

Please let me know if you agree.

The total 3 digit numbers are 999 including preceding zeros and there are 999/2 even total three digit even numbers are 499.

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