Combinations of $6$-digit natural numbers In each of the following 6-digit natural numbers: $333333,225522,118818,707099$,
every digit in the number appears at least twice. Find the number of such 6-digit natural numbers.
This is how I'm intending to do.
1) Find the total number of 6-digit combinations.
2) Subtract the number of times there's 0 and 1 repetition of digits so that I can get at least 2 repetitions. 
Total Combinations = $9 \times 10^5$
Repeat 0 Times = $9 \times 9 \times 8 \times 7 \times 6 \times 5$  
Repeat 1 Time = Not sure how to calculate
The answer is 11754  but I'm struggling to get it!
 A: First let us also accept numbers starting with a $0$.
Let $d$ denote the number of distinct digits in the number. For $d>3$ there are $0$ possibilities.
For $d=1$ there are $10$ choices of the digit and every choice leads to $1$ possibility. 
For $d=3$ there are $\binom{10}{3}=120$
choices for the digits and each choice leads to $\frac{6!}{2!2!2!}=90$
possibilities. 
For $d=2$ we have two split-ups. 
One of them is $6=3+3$
with $\binom{10}{2}=45$ choices for the digits and each choice leads
to $\binom{6}{3}=20$ possibilities. 
The other is $6=4+2$. Here the
chosen digits are distinghuisable. One of them is used $4$ times
and the other $2$ times. So we have $10\times9=90$ choices and each
choice leads to $\binom{6}{2}=\binom{6}{4}=15$ possibilities.
Adding up we find $10\times 1+120\times90+45\times20+90\times15=13060$ possibilities.

Subtracting the numbers that start with a $0$ we find $\frac{9}{10}\times13060=11754$
  possibilities.

A: I think I'd suggest direct counting, with cases being: (1) same digit six times; (2) a quadruple digit and a double digit; (3) two triple digits; (4) three double digits. Things get a bit messier because if one of your digits is $0$, you need to keep it out of the leading position.
Let's try counting case (3), and I'll leave the other cases to you:
Case 3a: Two triple digits, neither being $0$: Choose two nonzero digits ($9 \choose{2}$ ways).  Pick three of six positions for one of the digits ($6\choose 3$ ways). This gives a total of $36\cdot 20=720$ possibilities for this subcase.
Case 3b: Two triple digits, one being $0$: Choose a nonzero digit for the other digit (9 ways).  Choose $3$ of $5$ positions for the $0$s (you have to avoid the first position). This gives $9\cdot {5\choose 3}=90$ possibilities for this subcase.  
So your total for case 3 is $720+90=810$.
