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The problem goes so : you have a parking lot with 8 parking spaces and 8 cars, of which 4 are red and 4 are white. What is the probability of :

a) 4 white cars being parked next to each other ?

b) 4 white cars and 4 red cars being parked next to each other ?

c) red and white cars being parked alternately ( red-white-red...) ?

Any help will be greatly appreciated. :-)

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    Any thoughts? How many patterns are there in total? How many have, say, all four reds next to each other? – 2017-01-31
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    @lulu for the problem mentioned in a) I'd say there are 4 possible ways that white cars can be arranged so that they are parked next to each other – 2017-01-31
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    Why four? List them. – 2017-01-31
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    @lulu W-W-W-W-R-R-R-R, R-W-W-W-W-R-R-R, R-R-W-W-W-W-R-R, R-R-R-W-W-W-W-R,R-R-R-R-W-W-W-W, okay I miscalculated, there are 5 ways. – 2017-01-31
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    Right. And how many patterns are there in total? – 2017-01-31
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    @lulu there should be 8! ? – 2017-01-31
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    Well, there are $8!$ if you can distinguish all the cars. You could do it that way if you wanted, but then there are a lot more than $5$ ways to put all the white ones together . I think it's easier to keep the cars indistinguishable except for color. – 2017-01-31
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    @lulu how would I do that ? – 2017-01-31
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    There are eight slots, and you need to choose four (to place the white ones). – 2017-01-31
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    @ so it would be binom{8}{4} and so there would 5/ binom{8}{4} chance of a) happening, and for b) there would be 2/binom{8}{4} – 2017-01-31
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    Yes, that's right. – 2017-01-31

2 Answers 2

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Hint -

a.) Take 4 white cars as 1 group and arrange.

b.) Take 4 white cars as 1 group and 4 red cars as 1 group and arrange.

c.) Once start with red and arrange alternately and then start with white and arrange alternately. Combine both arrangements.

Take care that one type of cars are identical or distinguish.

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As you are just worried about color, it's good to solve this problem thinking in sets. What you really need as total cases is to choose 4 spaces from 8, which will corresponds to white (or red if you want, but not any), so $\binom{8}{4}$

a) Thinking in same way you have 5 favorable cases $\lbrace 1-4,2-5,3-6,4-7,5-8\rbrace$, so $5/\binom{8}{4}$; B) is the same.

c)You have two options: red evens or white evens, so $2/\binom{8}{4}$