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    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
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    (Original post by Zacken)
    I would think it would be 2 * 8! / (3! 3!) at a quick glance.
    thanks, why did you do that ?
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    (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)
    ...
    :holmes:

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

    Since cars of same colour are identical, total number of arrangements =\dfrac{8!}{2!3!3!}

    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" = \dfrac{7!}{3!3!}
    BUT: No need to multiply by two for there being two yellow cars as they are identical.

    So, #arrangements = =\dfrac{8!}{2!3!3!}-\dfrac{7!}{3!3!}=420
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    (Original post by ghostwalker)
    :holmes:

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

    Since cars of same colour are identical, total number of arrangements =\dfrac{8!}{2!3!3!}

    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" = \dfrac{7!}{3!3!}
    BUT: No need to multipy by two for there being two yellow cars as they are identical.

    So, #arrangements = =\dfrac{8!}{2!3!3!}-\dfrac{7!}{3!3!}=420
    thanks mate, makes sense )
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    I really thought this thread was going to be about hair
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    (Original post by ghostwalker)
    :holmes:
    Have deleted my posts. Thanks!

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