The answer is 2.

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I prefer the second way. For the number to be even, it is necessary and sufficient for the last digit to be 2.

oh how did you get it?

Pjhay Nedamla
- 7 years, 9 months ago

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.

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- xx2 ----> 2! = $\boxed{2}$

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X1 X2 X3 is the three digit no
X1- can have 3 values 1,2 or 3.
X2- can have 3 values 1,2 or 3.
but where as the term to be even last position X3 is occupied
only by 2.
so total no of digits are 3
*
3
*
1=9
so 9 is the answer

Bhaskar Akash
- 7 years, 3 months ago

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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 $\fbox{2}$

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sorry :P that should be $312$ instead of 321

JohnDonnie Celestre
- 7 years, 9 months ago

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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.