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A nine-digit number is formed using the digit $1,2,3,4,5,6,7,8,9.$ Find the probability of forming a number such that product of any of its $5$ consecutive digits is divisible by $3$ or $5$

What I tried

$A:$ events in which product of any $5$ consecutive digit is divisible by $3$

$B:$ events in which product of any $5$ consecutive digit is divisible by $5$

$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$

let any $5$ consecutive digits be $abcde$, where $a,b,c,d,e\in \{1,2,3,\cdots,9\}$

How do I solve this?

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For a set of $5$ numbers to have a product that divides either $3$ or $5$, the set must contain at least one member of the set $S=\{3,5,6,9\}$.

So we want to consider permutations of $9$ digits where any $5$ consecutive digits contains an element of $S$.

In other words, we can't have a permutation with a $5$ digit sequence without any elements of $S$. But, there are only $5$ such elements.

So, to count "bad" permutations, we can consider $V = \{1,2,4,7,8\}$, as there must be a section of these numbers consecutively. There are $5!$ permutations of $V$, and then $4!$ permutations of $S$, and $5$ places we can place the block.

So, our final answer is $$1-\frac{5\cdot5!\cdot4!}{9!}$$$$1-\frac{5!5!}{9!}$$$$1-\frac 5{126}=\color{red}{\frac{121}{126}}$$

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