How many five-digit numbers do not have three consecutive digits the same?

Question

How many five-digit numbers do not have three consecutive digits the same? Also, the initial digits might be $0$, but I'm not sure how that changes the answer.

This is the formula I've come up with for solving this problem. Total number of numbers - $A - B - C + A \cap B + B \cap C + A \cap C - A \cap B \cap C$ $$10^5 - (10^3 \cdot 3) + (10^2 \cdot 2) - 10$$

So I have set $A$, positions 1 2 3 are filled; set $B$, positions 2 3 4 are filled; set $C$, positions 3 4 5 are filled; so each set will have $10 \cdot 10 \cdot 10$ subsets of numbers that have $3$ consecutive numbers.

I know that their should be double counting because having consecutive numbers in positions 1 2 3 4 and 2 3 4 5 should be added back, and consecutive numbers 1 2 3 4 5 will need to be subtracted.

So, I get $$10^5 - (10^3 \cdot 3) + (10^2 \cdot 2) - 10 = 100000 - 3000 + 200 - 10 = 97190$$

However, this is not the correct answer. What is the correct procedure to solve this problem? Or what way am I to look at counting up the sets?

Thanks

Answer 1

There's one mistake and one possible misinterpretation.

The mistake is that you only substracted $10^2$ twice for $|A\cap B|$ and $|B\cap C|$, but $|A\cap C|$ is missing. This is actually the same as $|A\cap B\cap C|$, since in both cases all five digits have to be equal; so those two cancel and the correct total would be $10^5-3\cdot10^3+2\cdot10^2+10^1-10^1=97200$.

The potential misinterpretation is that the problem may have meant only "proper" five-digit numbers that don't start with a $0$.

Answer 2

I would do simple subtraction, viz.

all "numbers" - [all 5 digits same + 4 consecutive digits same + 3 consecutive digits same]

e,g, with consecutive 0's, the patterns can be: 00000 , x0000, 0000x, 000xx , x000x and xx000

$10^5 - [ 10 + 2\cdot10\cdot9 + 10\cdot9^2 + 2\cdot 10\cdot9\cdot10 ] = 97200$