#### williamrobertsuk

##### New member

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- Thread starter williamrobertsuk
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"At most one" means none or one. If they are parked side-by-side that can be done in \(\displaystyle n-1\) ways. WHY?

If they are parked with exactly one space between them that can be done in \(\displaystyle n-2\) ways. WHY is that?

What is the total number of ways the two cars can be parked in \(\displaystyle n\ge 2\) places?

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Thanks Mark, I am new at the forum, I will try to be better! I just try to find some results on those on some hobby I have!

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pka - It doesn't seem to be correct!

If they are parked with exactly one space between them that can be done in \(\displaystyle n-2\) ways. WHY is that?

What is the total number of ways the two cars can be parked in \(\displaystyle n\ge 2\) places?

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pka - It doesn't seem to be correct!

If they do not seem to be correct you must

Here is the model: Draw n empty boxes a page. Number each box \(\displaystyle 1,2,\cdots,n\).

If we randomly assign(by a draw, by a computer program ), two of those numbers to the two people. What is the probability their numbers will differ by no more than two? That mean the numbers are consecutive(next to each other) or there is one number between them(one space between them).

If that model is incorrect, then you must correct it. It is what you posted.

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Therefore, I am also a bit confused with it!Why did you ignore my request for reasons for the answers I gave?

If they do not seem to be correct you musttell me why they seem to be incorrect.

Here is the model: Draw n empty boxes a page. Number each box \(\displaystyle 1,2,\cdots,n\).

If we randomly assign(by a draw, by a computer program ), two of those numbers to the two people. What is the probability their numbers will differ by no more than two? That mean the numbers are consecutive(next to each other) or there is one number between them(one space between them).

If that model is incorrect, then you must correct it. It is what you posted.

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Tell us about your confusion. We cannot help if all you say is "I am confused"Therefore, I am also a bit confused with it!

Take the ten digits \(\displaystyle 0,1,2,3,4,5,6,7,8,9\) on balls as in the lottery. The machine picks two of them at random.

What is the probability the two numbers a consecutive or differ by one?

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So, here is the answer:

(4*n-6)/(n*(n-1))

(4*n-6)/(n*(n-1))

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The first car has 2 choices of spots where the second car has 2 choices (this is on either end, spots 1 and \(n\)).

The first car has 2 choices of spots where the second car has 3 choices (spots 2 and \(n-1\)).

The first car has \(n-4\) choices of spots where the second car has 4 choices (spots 3 through \(n-2\)).

Thus, adding these up, we find:

\(\displaystyle N_F=2\cdot2+2\cdot3+(n-4)4=4n-6\)

Hence:

\(\displaystyle P(X)=\frac{N_F}{N_T}=\frac{4n-6}{n(n-1)}\quad\checkmark\)