Question on combinatorics.. Please help !

Question

praxer · Answer

How many different seven digits number are such that the sum of the digit is even ?

kinggeorge · Answer

Presumably the first digit can't be zero? Or is that allowed?

praxer · Answer

No, it is not allowed.

kinggeorge · Answer

So this question is actually easier than it sounds. The first step, is to simply the choose the first 6 digits (all but the units digit). Can you tell me how many options you have for this?

praxer · Answer

9!   since the numbers can repeat... Am I correct ? :)

praxer · Answer

I think it should be 6*9!...

kinggeorge · Answer

Not quite. Remember that you have 9 options for the first digit (1-9). After that though, you have 10 options each for the 2nd through 6th digit. There are 5 digits there. So there are$$9\cdot10^5$$possibilities for the first 6 digits.

kinggeorge · Answer

Make sense so far?

praxer · Answer

understood it ...... :) then ?

kinggeorge · Answer

Now we just have the last digit. Now, if we add up the digits that we already have, it's either even or odd.

If the sum is even, the last digit must be 0,2,4,6, or 8. Anything else, and the sum of all the digits won't be even. If the sum is odd, the last digit must be 1,3,5,7, or 9.

There are 5 options for each, and so 5 options for the last digit. No matter what the sum of the previous digits is. We conclude that there are $$9\cdot10^5\cdot5$$7 digit numbers such that the sum of the digits is even.

praxer · Answer

so, it is $$ 45.10^5 $$... Thank you !

kinggeorge · Answer

You're welcome.