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!
Best Answer
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.