TITLE: How many 6 digit numbers are possible with no digit appearing more than thrice?
QUESTION [6 upvotes]: How many 6 digit numbers are possible with at most three digits repeated?
My attempt:
The possibilities are: 
A)(3,2,1) One set of three repeated digit, another set of two repeated digit and another digit (Like, 353325, 126161)
B)(3,1,1,1) One set of three repeated digit, and three different digits.(Like 446394, 888764)
C)(2,2,1,1) Two sets of two repeated digits and two different digits (Like, 363615, 445598)
D)(2,2,2) Three sets of two repeated digits (Like, 223344, 547547)
E)(2,1,1,1,1,1) One set of two repeated digit and four different digits (Like 317653, 770986)
F)(1,1,1,1,1,1) Six Different digits (like 457326, 912568)
G)(3,3) Two pairs of three repeated digits.
Let's try to calculate each possibilities separately.  
F) is the easiest calculate. 
Let us try to workout  Case E)
Let's divide the case into two parts:
Case E(1) Zero is not one of the digit 
We can choose any $5$ numbers form $9$ numbers $(1,2,3,\cdot, 9)$ in $\binom{9}{5}$ ways , the digit which one is repeated can be chosen in 5 ways, and you can permute the digits in $\frac{6!}{2!}$ ways. The total number of ways$=\binom{9}{5}\times 5\times \frac{6!}{2!} $
Case E(2) Zero is one of the digit.
Case E(2)(a) Zero is the repeated digit We need to choose four other numbers which can be done in $\binom{9}{4}$ ways, the digits can be permuted in $\frac{6!}{2!}$ ways, but we need to exclude the once which starts with zero ($5!$ many). The total number of ways =$=\binom{9}{4}\times  (\frac{6!}{2!} -5!)$.
Case E(2)(b) Zero is not the repeated digit We need to choose four other numbers which can be done in $\binom{9}{4}$ ways, the repeated digit can be chosen in 4 ways,  the digits can be permuted in $\frac{6!}{2!}$ ways, but we need to exclude the once which starts with zero ($5!$ many). The total number of ways =$=\binom{9}{4}\times 4\times  (\frac{6!}{2!} -5!)$.
Before, proceeding to workout the other cases, I want to know

Is my attempt correct?
If it is correct, it is too lengthy, is there any other way to solve this?

REPLY [3 votes]: I will assume that the question means that no digit appears more than three times.  As Austin Mohr points out, the question is badly phrased.
Since the leading digit cannot be zero, there are $9 \cdot 10^5 = 900,000$ six digit positive integers.  Like Soke, I will exclude those in which a digit appears four or more times.  
We consider cases.
Case 1:  The same digit is used six times.
Since the leading digit cannot be zero, there are $9$ of these.  They are $$111 111, 222 222, 333 333, 444 444, 555 555, 666 666, 777777, 888 888, 999 999$$
Case 2:  One digit is used five times, while a different digit is used once.
There are two subcases.
Subcase 1:  The leading digit is repeated.  
Since the leading digit cannot be zero, there are nine ways to select the leading digit.  We must select four of the remaining five places to place the other occurrences of the leading digit.  We then have nine choices for the other digit since we can now use zero.
$$9 \cdot \binom{5}{4} \cdot 9 = 405$$
Subcase 2:  The leading digit is not repeated.
We still have nine ways of selecting the leading digit.  That leaves us with nine ways to choose the repeated digit that fills the remaining five places.
$$9 \cdot 9 = 81$$
Case 3:  One digit is used four times, while a different digit is used twice.
Subcase 1:  The leading digit appears four times.
We have nine ways of selecting the leading digit.  We have $\binom{5}{3}$ ways of choosing the other three positions in which it appears.  We have nine ways of choosing the digit that fills the two open positions.
$$9 \cdot \binom{5}{3} \cdot 9 = 810$$
Subcase 2:  The leading digit appears twice.
We have nine ways of selecting the leading digit and $\binom{5}{1}$ ways of choosing the other position in which it appears.  We have nine choices for choosing the repeated digit that fills the four open positions.
$$9 \cdot \binom{5}{1} \cdot 9 = 405$$
Case 4: One digit is used four times, while two other digits are used once each.
Subcase 1:  The leading digit is repeated.
We have nine ways of choosing the leading digit and $\binom{5}{3}$ ways of choosing the other three positions in which it appears.  We have nine choices for the leftmost open position and eight choices for the remaining position.
$$9 \cdot \binom{5}{3} \cdot 9 \cdot 8 = 6480$$
Subcase 2:  The leading digit is not repeated.
We have nine ways of choosing the leading digit.  We have nine ways of choosing the repeated digit and $\binom{5}{4}$ ways of selecting four of the five open positions in which to place it.  We have eight ways of filling the remaining open position.
$$9 \cdot 9 \cdot \binom{5}{4} \cdot 8 = 3240$$
That gives a total of 
$$9 + 405 + 81 + 810 + 405 + 6480 + 3240 = 11,430$$
excluded cases.  
Hence, there are 
$$900,000 - 11,430 = 888,570$$
six-digit positive integers in which no digit appears more than three times.