Geomtric series - Finding the formula

[8-ball]

OK I got the following problem where I need to find a formula to determine the number of possible outcomes given N number of "cables".

The problem goes this way, I have three cables on of them is defective and I need to check them one at a time. If I start by inspecting cable 1 it can be either defective (first outcome) or non defective, if it is non defective then I cna either inspect cable 2 or cable 3. If I choose to inspect cable 2 it can be either defective (second outcome) or non defective which means that the third cable is defective (third outcome). If I choose to inspect cable 3 first instead of cable two then I have two more possible outcomes. 5 total outcomes if I start by inspecting cable 1, but I can start with cable 2 or cable 3.

So in total I have 15 possible outcomes if There are 3 Cables
Similarly I have 4 outcomes if there are 2 cables.
Similarly I have 64 outcomes if there are 4 cables.
Similarly I have 325 outcomes for 5 cables

So I need to develop a formula for N number of cables.
I believe that this should be in a form of a geometric series but I am not sure. This is what I have so far.

3 cables = (4 x 3) + 3 = 15
4 cables = (15 x 4) + 4 = 64
5 cables = (64 x 5) + 5 = 325
(previous number of cables x number of cables) + number of cables.

 

[DrMike]

(previous number of cables x number of cables) + number of cables.

This looks right to me. You could write this as :

\(\displaystyle O_n = nO_{n-1}+n\)

or

\(\displaystyle O_n = n\left(O_{n-1}+1\right)\)

As it stands, it's hard to solve. I'd suggest letting

\(\displaystyle O_n=n!P_n\)

Then,

\(\displaystyle O_{n-1}=(n-1)!P_{n-1}\)

and you get

\(\displaystyle n!P_n = n.(n-1)!P_{n-1}+n\)

that is,

\(\displaystyle P_n = P_{n-1}+1/(n-1)!\)

So,

\(\displaystyle P_n=P_0+\sum_{k=0}^{n-1}\frac{1}{k!}\)

 

[8-ball]

(previous number of cables x number of cables) + number of cables.

\(\displaystyle O_n = n\left(O_{n-1}+1\right)\)

As it stands, it's hard to solve. I'd suggest letting

\(\displaystyle O_n=n!P_n\)


\(\displaystyle P_n=P_0+\sum_{k=0}^{n-1}\frac{1}{k!}\)

I am sorry, I got lost when you introduced the first

\(\displaystyle P_n\) Also, I made I mistake on my formula, It shoule be

Number of outcomes for N cables = (previous number of outcomes for N-1 cables x number of cables) + number of cables