1, 2, 3

Well there are two major ways of doing this problem. One is by listing out and finding out all even numbers comprising only these three numbers. We know that these digits do not repeat.

So: 123, 132, 213, 231, 312, 321.

Essentially, we count these and the answer we see is 2.

Another way is that since only three digits will be used without repeating digits and that it has to be even, the ones digit has to be an even integer since that is the only way any number can be presumed as even. So we 2 as our only even integer so that leaves 3 and 1 for 2 places. We have 2 times 1 which is 2 .

There are 6 combinations for 3-digit numbers containing the digits 1, 2, and 3. They are:

123 132 231 213 312 321

Of these, the only combinations that are even are:

132 312

Therefore, there are 2 even 3-digit numbers containing the digits 1, 2 and 3.

There 6 ways to form the numbers 1, 2 and 3 into a three-digit number: 123, 132, 213, 231, 312, and 321. Of these numbers, there are only two even numbers, 132 and 312. Therefore the answer is 2.

Since we want to get even numbers, 2 should be the unit digit. Thus, we are left with 1 and 3, which yields 2 permutations, i.e. (1,3) and (3,1). In conclusion, there are only 2 numbers, 132 and 312.

• xx2 ----> 2! = $\boxed{2}$

132 and 312

There are 3! three-digit numbers which can be formed from the digits 1, 2, and 3. They are: 123, 132, 213, 231, 312, 321. Of these, the only even ones are 132 and 312. So the answer to this question is 2.

From the given digits $1, 2, 3$ , a 3-digit number is formed if and only if 2 is its last digit.

This restricts us that $1$ or $3$ takes the hundredths or tenths place, which gives $132$ and $321$ . So there are $\fbox{2}$

there are six 3 digit number can be form from 1 2 3 321 312 231 213 123 132 but 312 and 132 only are even .