2011 AMC 10A Problems/Problem 13
Problem 13
How many even integers are there between 
and 
whose digits are all different and come from the set 
?
Solution 1
We split up into cases of the hundreds digits being 
or 
. If the hundred digits is , then the units digits must be 
in order for the number to be even and then there are 
remaining choices (
) for the tens digit, giving 
possibilities. Similarly, there are 
possibilities for the case, giving a total of 
possibilities.
Solution 2
We see that the last digit of the 
-digit number must be even to have an even number. Therefore, the last digit must either be or .
Case 
-the last digit is . We must have the hundreds digit to be and the tens digit to be any of 
, thus obtaining numbers total.
Case -the last digit is . We now can have or to be the hundreds digit, and any choice still gives us choices for the tens digit. Therefore, we have 
numbers.
Adding up our cases, we have 
numbers.
Solution 3 (elimination)
We see that there are 
total possibilities for a 3-digit number whose digits do not repeat and are comprised of digits only from the set 
Obviously, some of these (such as 
for example) will not work, and thus the answer will be less than 
The only possible option is 
~ Technodoggo
- This doesn't make sense because there are
to make a 3-digit number from the given set of 6 numbers.
to make a 3-digit number from the given set of 6 numbers.See Also
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