What is wrong in my solution ?
Q: If we have $N$ number of light leds, and the probability of having faulty ones is $R$ (in percentage), what is the probability of having 2 or more faulty light leds next to each other?
My solution: using permutation:
$$ P=\left(\frac{2! \, (N -1)!}{N!}\right) R $$
Any useful links for similar cases is appreciated too.
Thank you
The case of two:
You choose the two positions with $n-1$ ways.
Then you can choose two positions with $\binom{n}{2}$ ways.
If the leds are discrite and only then, you do *:
*Then you permute them with $2!$ ways.
*Then you permute the other elements with $(n-2)!$ ways.
Multiply the possibility of finding two neighbour positions with $R$.
Continue for the case of $3$ until the case of $n-1$.
Sum up!