Thread: combinatorics problem involving even numbers?

The problem is: How many even numbers greater than 40,000 may be formed using the digits 3, 4, 5, 6 and 9 if each digit must be used exactly once in each number?

Like all combinatorics problems, I create five spots. The last spot must be used up by 4 or 6, b/c it's even, so we have

_ _ _ _ 2

But then my book says to assign the first spot to 3, because there are 3 numbers left over after choosing two for the last spot. But since I only used one number for the last spot--either 4 or 6---shouldn't the first spot be assigned to 4, and then, accordingly:

4 3 2 1 3 ?

Originally Posted by lu6cifer

The problem is: How many even numbers greater than 40,000 may be formed using the digits 3, 4, 5, 6 and 9 if each digit must be used exactly once in each number?

Like all combinatorics problems, I create five spots. The last spot must be used up by 4 or 6, b/c it's even, so we have

_ _ _ _ 2

But then my book says to assign the first spot to 3, because there are 3 numbers left over after choosing two for the last spot. But since I only used one number for the last spot--either 4 or 6---shouldn't the first spot be assigned to 4, and then, accordingly:

4 3 2 1 3 ?

The key here is "greater than 40'000" - the first number CANNOT be that 3. Thus we have 3 choices for the first number. So we have,

3 3 2 1 2

which gives us 36 choices.

ooo..you're right

must've forgotten that part amidst all the combinatorics stuff.