# How many six digit numbers have $3$ even and $3$ odd digits? [closed]

How many six digit numbers have $3$ even and $3$ odd digits?

It would be really helpful it someone could boil down to the absolute basics and explain me how I can solve this!

## closed as off-topic by Did, user91500, Namaste, kingW3, Sahiba AroraAug 19 '17 at 15:20

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• I know there is a similar question which involves a $20$ digit number but I couldn't exactly understand the solution! – MathDude3013 Aug 19 '17 at 8:03
• I tried taking 2 cases and then permuting them! Besides this, there is some $5^n$ stuff which I am not getting – MathDude3013 Aug 19 '17 at 8:05
• Do your two cases involve whether the leading digit is even or odd? – N. F. Taussig Aug 19 '17 at 8:08
• @N.F.Taussig yes! – MathDude3013 Aug 19 '17 at 8:09
• There is some $5^n$ stuff in the solution which I am not getting! – MathDude3013 Aug 19 '17 at 8:11

Since the leading digit cannot be zero, we break the problem into two cases, depending on whether the leading digit is even or odd.

Leading digit is even: Since the leading digit cannot be zero, we can fill the hundred thousands place in four ways, namely with $2$, $4$, $6$, or $8$. We choose two of the remaining five positions for the even digits. Each of those positions can be filled in five ways, namely with one of the five even digits. Each of the three remaining positions can also be filled in five ways, namely with one of the five odd digits.

$$4\binom{5}{2}5^5$$

Leading digit is odd: We choose two of the remaining five positions for the odd digits. There are five ways to fill each of the six positions.

$$\binom{5}{2}5^6$$

Total: Since the two cases above are mutually exclusive and exhaustive, the total number of six-digit positive integers with three even and three odd digits is found by adding the above results.

The even digits are $0, 2,4,6,8$ and odds are $1,3,5,7,9$. Now, We have six place which can be fill like the following:

• first choose 3 places from six places to put even numbers: $\binom{6}{3}$
• then you can place the digits in $5^6$ different combination (there are six places and you have 5 options for each place). So, all combination is $\binom{6}{3}\times 5^6 -$ #number of placed zero as the first digit. Therefore the final result is: $$\binom{6}{3}\times 5^6 - \binom{5}{2}5^5 = 90\times 5^5 = 18\times 5^6$$