First, the computed result (code below) is 6269585184

Analytic estimate

The total ways to choose such a number is 5*6*6*6*6*3 = 19440, but then we must subtract the number of these which have no repeat digits.

That would be 5*5*4*3*2*1 = 600 if evens and odds were both allowed, so 300 of these would be even.

We have 19440 - 300 = 19140 ways.

Determine, for each digit position what the average digit size is.

Do this if no repeat digits is allowed; and also for the no repeat digits group.

1st group (allowing "all unique digits") average digit going from left to right:

    3, 2.5, 2.5, 2.5, 2.5, 2

    Multiplying each of these by the appropriate power of 10 and summing: 327777

    This would be the average size of N (if no repeat digits were allowed)

A similar analysis for six digit pandigitals (0...5) which are even hits a snag because constraining the last digit to be even, throws off the probabilities for the other digits: they are less likely to be even than they would be if the "even" constraint was not there.  One could figure it out without the even constraint, then divide by 2 and this would be pretty close, so lets do that.

2nd group:  6 digit pandigitals (0...5), ie all unique digits, including evens AND odds, average digits are:

    3, 2.5, 2.5, 2.5, 2.5, 2.5

Now the average group 2 number is 327777.5 and there would be 600 of them (5*5*4*3*2*1) if evens and odds were both allowed.   So 327777.5 * 300 should be close:  98333250

Multiplying 327777 (which is not accurate) by 19440 and then subtracting 327777.5 * 300 should give a rough estimate of the answer:  

6273651630 is the rough estimate

6269585184 is the computed answer

Sources for error in the rough estimate:

    Constraining the last digit to be even tweaks the probabilities of being even for the first 5 digits.

    The sum of all even 6-digit pandigitals (0...5) is probably not exactly 1/2 of the sum of all the evens plus the odds.

Summary:  1.5 cheers for an analytic estimate in the right ballpark, but acknowledge a failure of obtaining an accurate analytic solution.

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bigsum = 0

for a in range(1,6):

    for b in range(0,6):

        for c in range(0,6):

            for d in range(0,6):

                for e in range(0,6):

                    for f in range(0,6,2):

                        if len(set([a,b,c,d,e,f])) == 6:

                            continue

                        littlesum = f+e*10+d*100+c*1000+b*10000+a*100000

                        bigsum += littlesum

print(bigsum)