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# perms and combs Watch

1. help w/ this question

1. Eight cars (3 red, 3 blue, and 2 yellow) are to be parked in a line.How many unique lines can be formed if the yellow cars must notbe together? Assume that cars of each color are identical.

I've done 8!-(7!x2!) / 3! =5040

but not sure if its right
2. (Original post by Zacken)
I would think it would be 2 * 8! / (3! 3!) at a quick glance.
thanks, why did you do that ?
3. (Original post by torifoster97)
1. Eight cars (3 red, 3 blue, and 2 yellow) are to be parked in a line.How many unique lines can be formed if the yellow cars must notbe together? Assume that cars of each color are identical.
2. I've done 8!-(7!x2!) / 3! =5040
(Original post by Zacken)
...

#arrangements = "total #arrangements" - "#arrangements where both yellow cars are together".

Since cars of same colour are identical, total number of arrangements

If both yellow cars are together we can treat them as one item, as Zacken did, to get:

"#arrangements where both yellow cars are together" =
BUT: No need to multiply by two for there being two yellow cars as they are identical.

So, #arrangements =
4. (Original post by ghostwalker)

#arrangements = "total #arrangements" - "#arrangements where both yellow cars are together".

Since cars of same colour are identical, total number of arrangements

If both yellow cars are together we can treat them as one item, as Zacken did, to get:

"#arrangements where both yellow cars are together" =
BUT: No need to multipy by two for there being two yellow cars as they are identical.

So, #arrangements =
thanks mate, makes sense )
5. I really thought this thread was going to be about hair
6. (Original post by ghostwalker)
Have deleted my posts. Thanks!

(I am not awake enough for this, as can be seen. )

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