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Thread: combination/permutation help!!!!! pls!!

  1. #1
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    combination/permutation help!!!!! pls!!

    I have a big question about permutation here. the first question, I can understand the way how to solve. but I used the same way to solve the 2nd question, it was wrong, please help me here!!! why I can't use the same way to solve the 2nd one??
    (practice for gmat!)



    question 1: 8 cars (3red, 3blue, 2 yellow)are to be parked in a line, how many unique lines can be formed if the yellow cars must not be together? assume that cars of each colors are identical.

    8!-2!*7!=18

    I know how to solve this one. but 2nd one is harder!

    question 2: how many ways can 8 books, each covering a different subject, be arranged on a shelf such that books on biology, history, or programming are never together?

    8!-3!*5!= this is wrong, but I don't get it!

    pls if you could help, I do need some help. thanks!
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  2. #2
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    Re: combination/permutation help!!!!! pls!!

    Your 5 should be a 6.
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    Re: combination/permutation help!!!!! pls!!

    yeah, I made a mistake. it was 6. but 8!-3!*6! is still wrong.
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    Re: combination/permutation help!!!!! pls!!

    Ok. You'll also need to take into account when biol and history are together but not with programming, etc. Eg * * * B H * * P .
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    Re: combination/permutation help!!!!! pls!!

    Quote Originally Posted by fightmath View Post
    question 1: 8 cars (3red, 3blue, 2 yellow)are to be parked in a line, how many unique lines can be formed if the yellow cars must not be together? assume that cars of each colors are identical.
    8!-2!*7!=18 INCORRECT

    question 2: how many ways can 8 books, each covering a different subject, be arranged on a shelf such that books on biology, history, or programming are never together?
    8!-3!*5!= this is wrong, but I don't get it!
    #1 there are $\dfrac{6!}{(3!)^2}=20$ ways to arrange the red & blue cars. They create seven places for the yellow cars.
    Final answer $\dfrac{6!}{(3!)^2}\cdot \dbinom{7}{2}=420$ ways
    Thanks from fightmath
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    Re: combination/permutation help!!!!! pls!!

    Let's try it your way.

    Why 8!. Ignoring the restriction against the two yellow cars being placed adjacently and putting signs on the red cars marked a, b, and c, on the blue cars marked d, e, and f, and on the yellow cars marked g and h, you can indeed form 8! distinguishable lines.

    Now remove the signs from the red cars. The sign marked a could be on the first red car from the left, or the second red car from the left, or the rightmost red car. That's 3 possibilities. That leaves 2 red cars that could have the sign marked b. So removing the signs from the red cars reduces the number of distinguishable lines by a factor of 6 = 3!. Now remove the signs from the blue cars. Another reduction by a factor of 3!. Now remove the signs from the yellow cars. A reduction by a factor of 2 = 2!. So the relevant number of distinguishable lines, ignoring the restriction on adjacently placed yellow cars, is

    $\dfrac{8!}{3! * 3! * 2!} = \dfrac{8 * 7 * 6 * 5 * 4 * 3 * 2}{3 * 2 * 3 * 2 * 2} = 8 * 7 * 5 * 2 = 560.$

    Now if you follow the same logic with respect to just red and blue cars, you get the number of distinguishable lines of cars as

    $\dfrac{6!}{3! * 3!} = 5 * 4 = 20.$

    (This by the way is Plato's starting point, but we are doing this your way.)Now let's think where we can put two yellow cars adjacent to one another.

    They could go in positions 1 and 2, or 2 and 3, or 3 and 4, etc for a total of 7 possibilities . So the possible arrangements with two adjacent yellow cars is

    $7 * 20 = 140$.

    Thus, the answer is $560 - 140 =420$ which Plato got by a slightly different method.
    Last edited by JeffM; May 20th 2016 at 08:24 AM.
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    Re: combination/permutation help!!!!! pls!!

    I didn't finish it. [8!-2!*7!]/3!3!2! =420, is this right?

    thanks for reminding me, otherwise I though that's the way to solve it!!
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    Re: combination/permutation help!!!!! pls!!

    Quote Originally Posted by JeffM View Post
    Let's try it your way.

    Why 8!. Ignoring the restriction against the two yellow cars being placed adjacently and putting signs on the red cars marked a, b, and c, on the blue cars marked d, e, and f, and on the yellow cars marked g and h, you can indeed form 8! distinguishable lines.

    Now remove the signs from the red cars. The sign marked a could be on the first red car from the left, or the second red car from the left, or the rightmost red car. That's 3 possibilities. That leaves 2 red cars that could have the sign marked b. So removing the signs from the red cars reduces the number of distinguishable lines by a factor of 6 = 3!. Now remove the signs from the blue cars. Another reduction by a factor of 3!. Now remove the signs from the yellow cars. A reduction by a factor of 2 = 2!. So the relevant number of distinguishable lines, ignoring the restriction on adjacently placed yellow cars, is

    $\dfrac{8!}{3! * 3! * 2!} = \dfrac{8 * 7 * 6 * 5 * 4 * 3 * 2}{3 * 2 * 3 * 2 * 2} = 8 * 7 * 5 * 2 = 560.$

    Now if you follow the same logic with respect to just red and blue cars, you get the number of distinguishable lines of cars as

    $\dfrac{6!}{3! * 3!} = 5 * 4 = 20.$

    (This by the way is Plato's starting point, but we are doing this your way.)Now let's think where we can put two yellow cars adjacent to one another.

    They could go in positions 1 and 2, or 2 and 3, or 3 and 4, etc for a total of 7 possibilities . So the possible arrangements with two adjacent yellow cars is

    $7 * 20 = 140$.

    Thus, the answer is $560 - 140 =420$ which Plato got by a slightly different method.
    thanks so much for helping me!!!
    I though I was right about this one.

    But what I can't solve is question 2. more than 2 things I have to consider, I don't know how to do it!!
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    Re: combination/permutation help!!!!! pls!!

    Quote Originally Posted by Plato View Post
    #1 there are $\dfrac{6!}{(3!)^2}=20$ ways to arrange the red & blue cars. They create seven places for the yellow cars.
    Final answer $\dfrac{6!}{(3!)^2}\cdot \dbinom{7}{2}=420$ ways
    I don't get it (7/2)?
    Last edited by fightmath; May 20th 2016 at 12:39 PM.
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    Re: combination/permutation help!!!!! pls!!

    Quote Originally Posted by Plato View Post
    #1 there are $\dfrac{6!}{(3!)^2}=20$ ways to arrange the red & blue cars. They create seven places for the yellow cars.
    Final answer $\dfrac{6!}{(3!)^2}\cdot \dbinom{7}{2}=420$ ways
    also, could you please tell me how to solve the 2nd question that I need to know? thanks!!
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    Re: combination/permutation help!!!!! pls!!

    Quote Originally Posted by fightmath View Post
    I don't get it (7/2)?

    question 2: how many ways can 8 books, each covering a different subject, be arranged on a shelf such that books on biology, history, or programming are never together?
    First of all IT IS NOT $\bf{\dfrac{7}{2}}~!$ Why are you doing these if you don't know about binomial coefficients?
    $\dbinom{N}{k}=\dfrac{N!}{k!\cdot(N-k)!}$.

    #2, The wording implies that each book is unique and we do not want any two books on the same subject together.
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    Re: combination/permutation help!!!!! pls!!

    Quote Originally Posted by Plato View Post
    First of all IT IS NOT $\bf{\dfrac{7}{2}}~!$ Why are you doing these if you don't know about binomial coefficients?
    $\dbinom{N}{k}=\dfrac{N!}{k!\cdot(N-k)!}$.

    #2, The wording implies that each book is unique and we do not want any two books on the same subject together.
    ok, I think I got it.

    but the 2nd question, I need to know how to solve it. are those two questions are the same things?
    I don't know how to analyze it.
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    Re: combination/permutation help!!!!! pls!!

    If I remember correctly, Plato taught the course that you are taking for many years. There are formulas and notation and concepts that you must get comfortable with if you are going to do well on timed tests.

    In addition, it is really important to understand the logic behind various formulas because otherwise you will use the wrong one.

    Let's start with an easier problem. There are eight books, one each on one of eight topics. Let's label the positions on the shelf from left to right as a, b, c and so on. How many ways can you slot books into positions. You have 8 choices for position a, 7 for position b, etc. So there are 8! possibilities.

    Now you are supposed to exclude cases where history and biology are immediately adjacent to each other. One such case is history in position a and biology in position b. Another is history in position d and biology in position e. A third is biology in position d and history in position e. How many possibilities in total for history and biology being adjacent? Is that total a number to subtract from 8!. NO NO NO. We have to account for the other six books for EACH case of paired history and biology. So the adjustment for paired history and biology is what?

    Now think about the other prohibited pairs. What do you think the adjustment is for them.

    Are we good now? How about triplets? What do we do about them?

    Now has there been any double counting along the way? If so, how do we adjust for it?

    Notice that there is nothing hard here about calculations. What is hard is organizing your thoughts in a systematic way.

    Give it a try on your own. I'll be back in a few hours, and perhaps Plato will be back sooner. (Plato REALLY likes to see work from the student, and I am quite fond of it myself. Neither of us will be there when you take the test.)
    Last edited by JeffM; May 20th 2016 at 04:01 PM.
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    Re: combination/permutation help!!!!! pls!!

    Quote Originally Posted by fightmath View Post
    ok, I think I got it.

    but the 2nd question, I need to know how to solve it. are those two questions are the same things?
    I don't know how to analyze it.
    Consider the letters B_1,~B_2,~B_3,~H_1,~H_2,~H_3. There are six unique letters.
    How many ways can we rearrange these six letters so that no two letters of the same type are together?
    Here is one way: H~B~H~B~H~B. Now that is without subscripts. There is only one other way. Why is it?
    There are $3!$ ways to assign the subscripts to the H's AND there are $3!$ ways to assign the subscripts to the B's.
    So the string B_1,~B_2,~B_3,~H_1,~H_2,~H_3 can be rearranged in $72$ ways so on two of the same type are together.

    We want to add the letters P_1~\&~P_2 to any one of those $72$ strings so that the two P's are not together.
    There are $7\cdot 6$ ways to do that. Explain why that is.

    So the string B_1,~B_2,~B_3,~H_1,~H_2,~H_3,~P_1,~P_2 can be rearrange so that no two letters of the same type are together in $72\cdot 42$ ways.

    (B) for biology; (H) for history; (P) for programming.
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