# How many 3 digit even numbers are there(No Repetition)?

First find numbers ending with 0

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?

• Why do you say "2 digits are already consumed"? Does the problem state that the digits must be distinct? – ShreevatsaR Feb 16 '14 at 10:55
• @ShreevatsaR Yes,Edited question title – techno Feb 16 '14 at 10:56
• (1) You need to consider other possible last digits too, not just 0. (2) You're saying that for the 100s place you can't have the other 2 digits, nor can you have 0. This is correct, but that's not necessarily 3 digits ruled out: it depends on whether 0 is one of the last 2 digits. – ShreevatsaR Feb 16 '14 at 11:02

First find the even numbers that are ending up in zero so, for no. ending up with zero are zero at one's place so 1 combination now 9 at hundreds and 8 at tens Now even numbers not ending with zero i.e. ending in 2,4,6,8 so 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 place so, total no. of digit sequences would be =>9*8*1+8*8*4=72+256=328 When 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.

• +1 :) this the only answer that matches with the solution in my book – techno Feb 16 '14 at 11:08
• Why do you go for the 100's place after filling the Unit's place?Rather than going to the ten's place? – techno Feb 16 '14 at 11:17

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 = 32 CASE C: If 0 is not there at all, No of terms possible is: 4x8x7 = 224

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

• There cannot be a zero in the hundreds place since you would then not have a three-digit number. It looks like you meant a zero in the tens place. – N. F. Taussig Nov 6 '15 at 11:32

\begin{align} & \underline{\text{case 1}}\text{ : All possible even 3-digit numbers are given by = {${5*9*8 =360}$} } \\ & \underline{\text{case 2}}\text{ : All even number with ''0'' at }hundreds\text{ }place\text{ are given by = {${4*8*1=32}$} } \\ & \underline{The\text{ number of even 3 digit is = case1 -case 2 = {360-32=328}} } \end{align}

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 0,2,4,6,8 So total 3 digits even numbers would be 9*10*5 =450

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.

On the number line the 3 digit numbers are : 100 - 999 So if I would start at 1 - 999 then I have 999 numbers in total. from those I will take away the one digit numbers : 9 and the two digit numbers: 90 I.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 numbers. so total three digit even numbers are 499.

• $100$ itself is a 3 digit number so i believe that there are $1000$ 3 digit numbers, not $999$ – Yiyuan Lee Feb 16 '14 at 11:08