# How many EVEN six-digit numerals have no repeating digits?"The Fundamental Counting Principle" question Consider a six-digit numeral.(Note 022713) is classified as the 5-digit numeral 22713) How...

How many EVEN six-digit numerals have no repeating digits?

"The Fundamental Counting Principle" question

Consider a six-digit numeral.(Note 022713) is classified as the 5-digit numeral 22713)

How many EVEN six-digit numerals have no repeating digits?

For this question, I am quite confused because it is asking for even numbers yet it consists of all 10 digits. I also don't understand why it tells me about a number such as 022713. There is also an odd number question, but I find this one more confuzzling. Please explain it as thorough as possible I really don't get this unit at all and I have an exam coming up >.<

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The question tells you about 022713 because the numbers you need to consider are going to be from 100000 to 999999. Also each digit may be any of 0 through 9 except for the ones digit, since the number must be even. This means the ones digit can only be one of 0, 2, 4, 6, 8.

To find out how many possible digits there are, consider a six digit number ABCDEF, where each letter will have some digit in it. We know that the number is even, so F can only be one of 0, 2, 4, 6, 8, which means there are 5 possibilities.

Now consider the digit for A. If F=0, then there are 9 possible values for A. If `F\ne 0`, then A can be any of the remaining 8 digits. Then for each of the remaining digits, we have to reduce the number of possible digits, since we can have no digits repeating. This means that we have two possibilities:

Case 1: `F=0`

`9 times 8 times 7 times 6 times 5 times 1=15120`

``where each number represents the number of possible values of that digit.

Case 2:

`8 times 7 times 6 times 5 times 4 times 5=33600`

So the total number of possible numbers is those from case 1 added to those from case 2. This means there are a total of `15120+33600=48720` numbers.