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
Best Answer
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 $0$.