We should form two cases, one of 0 (because 0 can't be the first one) and one with all numbers 1-9, to find out how many numbers there are that use a digit 3 or more times. Then, we can subtract this from 90,000 (the number of five digit numbers). This will give us the desired number.
Case 1: 0 shows up more than 3 times.
When you think about it, the only way 0 shows up more than three times is if the number is 10,000; 20,000; 30,000; etc., up to 90,000. So we just have to add nine to the following case.
Case 2: Some number 1-9 shows up more than 3 times.
This is slightly more tricky. We start with 1, just to make things easy. But when you really start to think about it, 1's would have to fill 4/5 slots in the number for it to break the rules. Therefore, the numbers (if the 5th number is 2) would be as follows:
11112
11121
11211
12111
21111
and we can do the same thing for every other number, 0 and 2-9. However, 01111 does not count, so we have to make sure to get rid of that one as well. Therefore, there are 8 times 5, or 40 ways this can be done for the number 1 for 2-9, and 4 ways this can be done with 0. Overall, the number 1 can be made in 44 different ways. This is also true for 2, 3, 4, 5, 6, 7, 8, and 9 (try plugging them in above where 1 is now). This gives us 9*44 + 9 ways to break the rules, or 9*45 ways. This comes out to 405 ways.
Subtract this from 90,000 and you will see that there are 89,595 ways to accomplish your task.
A start: Here it is easier to count the complementary number: the number of 6 digit positive integers with four or more of one digit repeated.
This reduces to only four cases: (4, 1, 1), (4, 2), (5, 1), (6).
The rest:
(6) is simply $9$. These are $(111111, 222222, \dots, 999999)$.
(5, 1) is a little trickier. First there are $9 * 8$ ways to choose two distinct digits that aren't either zero and assign each to $5$ or $1$. Then there are ${6 \choose 1}$ ways to permute the order.
Now there's case where one is zero. There are $9$ ways to choose the other digit (it can't be zero or else the entire number is zero). For each combination, there are similarly ${6 \choose 1}$ ways to permute the order. In particular, there are ${6 \choose 1}$ ways to permute five zeros and one of the other digit and another ${6 \choose 1}$ ways to permute one zero and five of the other digit. Exactly half of these are valid by symmetry.
Indeed, the bijection that flips each digit demonstrates this property: if we map, e.g., $aaaa0a \mapsto 0000a0$, exactly one of these will be valid for each pair. In this case, we have $a00000, a0aaaa, aa0aaa, aaa0aa, aaaaa0a, aaaaa0$ as valid strings. In total this gives
$$(9 * 8 + 9) * {6 \choose 1} = 486.$$
(4, 2) is similar. There are $9*8$ ways to initially choose and then ${6 \choose 2}$ ways to permute the order.
If one is zero, again there are $9$ ways to choose the other digit and for each combination the number of ways to permute the order is ${6 \choose 2}$. This is
$$(9 * 8 + 9) {6 \choose 2} = 1215.$$
Finally (4, 1, 1). There are $9 * {8 \choose 2}$ ways to choose non-zero digits and assign them to a frequency. There are then $6! / 4!$ permutations. This gives
$$9 * {8 \choose 2} * 6! / 4! = 7560.$$
Now choose a triplet of unique digits $(a, b, c)$ where one is zero. There are ${9 \choose 2}$ ways of doing so. Now, consider the ${6 \choose 4}$ ways to make a string from four $r$'s (r for repeated) and two $s$'s (s for single). If we are given $rsrrrs$, for example, there are now $3!$ ways to choose one of the digits as the repeated and the other two each into one $s$ spot. Of these, two are invalid, namely when we choose the repeated digit to be zero. You can convince yourself this holds for any string. Thus, this gives
$$4{9 \choose 2}{6 \choose 4} = 2160$$
The grand total is $11430$. There are $900000$ six digit numbers in total, so the desired number is $\boxed{888570}$.
Verified solution on computer in Python:
def get_frequencies(cycle):
result = 0
for num in range(10**5, 10**6):
digit_freq = [0]*10
for digit in get_digits(num):
digit_freq[digit] += 1
digit_cycle = sorted([x for x in digit_freq if x != 0], reverse=True)
if digit_cycle == cycle:
result += 1
return result
def get_digits(num):
r = []
while num > 0:
r.append(num % 10)
num /= 10
return r
Running with the following main function:
def main():
print get_frequencies([6], 6)
print get_frequencies([5, 1], 6)
print get_frequencies([4, 2], 6)
print get_frequencies([4, 1, 1], 6)
Outputs the following lines:
9
486
1215
9720
Or, more directly, we can use this program:
def get_frequency_atmost(max_freq, n):
result = 0
for num in range(10**(n-1), 10**n):
digit_freq = [0]*10
for digit in get_digits(num):
digit_freq[digit] += 1
if max(digit_freq) > max_freq:
result += 1
return 10**n - 10**(n-1) - result
which prints $888570$ on input get_frequency_atmost(3, 6).
Best Answer
The last digit can be determined by $5$ different ways. If it is $0$, then the two other digits are obtained in $9\times 8$ different cases.
If the last digit is something other than $0$, then the first digit can be determined by $8$ different cases (other than the last digit and $0$) and the middle one in $8$ different cases (excluding the first and last digits). Hence the total cases are $$9\times 8+4\times 8\times 8$$