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Thread: permutations&combinations

  1. #1 permutations&combinations 
    Forum Ph.D. Heinsbergrelatz's Avatar
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    hey everyone, im new to this website, and i would like to start up with a question that had been kind of confusing, and tedious

    the topic is on permutations and combinations;
    a)find the number of ways 3boys and 4 girls can be arranged in a row if
    1)there are no restrictions
    2)a particular boy and a particular girl must be next to each other
    3)the first and last persons are of the opposite sex


    and also how do you know what questions apply for Permutations, or combinations, how do you know which one to use? [/code]

    [/quote]


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  3. #2  
    Forum Freshman pbandjay's Avatar
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    I am confused on this problem. You have 3 boys / 4 girls that must be lined up such that no two boys and no two girls are next to each other. I only see this possible if you line them up like this:

    G B G B G B G

    For if you have a boy in the first spot, you get:

    B G B G B G G ; which is not possible because now two girls are lined up together.

    However, the first one cannot be correct because you said the first and last place must be of the opposite sex. If this is a textbook problem, are you sure you copied everything correctly? If so I think I must be missing something... (a good place to start with any combinatorics/counting problem is to actually draw out your line or pattern that you are trying to count).

    As far as your second question... I have to actually visualize the situation that is being counted to basically see whether the "order matters". In your example, you are forming a line of actual people. I think it is reasonable to assume that each person is a unique individual and Girl 1 is different from Girl 2, therefore you would want to count these two situations as separate:

    G1 B1 G2 B2 ...
    G2 B1 G1 B2 ...

    Another good example could be: suppose you are a professor passing out one piece of chalk to each student in your class. If each piece of chalk is a different color, then permutations would be used. Trading the pieces of chalk between the students renders different cases.

    However, if each piece of chalk was identical and of the same color, the trading would not mean anything--each student still has the "same" piece of chalk, so if you count with permutations, you over count (largely). In this case would divide by the number of ways it is possible to over count, or (more simply), you use combinations.


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  4. #3  
    Ots
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    I think it's a three-part question.

    Find the number of ways 3 boys and 4 girls can be arranged in a row if:

    1. There are no restrictions. So how many ways can 7 people be arranged in a line.
    The answer here is probably 7!

    Next,

    2. How many ways can they be arranged if a particular boy and a particular girl have to be next to each other (paired). I would see this as a pair, two boys and three girls or six 'things' so probably 6!

    And lastly,

    3. If opposite sexes are at the ends. If the ends of G...B and B...G are considered to be the same, then 5! Basically, how many ways can five people be arranged in between a boy and a girl.

    I think my way of arranging makes the girls and boys unique. So B1 is different from B2. If there is no difference then the formulas are different.
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  5. #4  
    Forum Freshman pbandjay's Avatar
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    Quote Originally Posted by Ots
    2. How many ways can they be arranged if a particular boy and a particular girl have to be next to each other (paired). I would see this as a pair, two boys and three girls or six 'things' so probably 6!
    I am not so sure about this. There is only a single way to actually arrange the boys/girls such that they are standing G B G B G B G. There are 4 spots for the girls, 4!. There are 3 spots for the guys, 3!. By the multiplication principle, the answer would be 4!3!. I just don't see how 6! limits the positions to boy girl boy girl...

    Quote Originally Posted by Ots
    3. If opposite sexes are at the ends. If the ends of G...B and B...G are considered to be the same, then 5! Basically, how many ways can five people be arranged in between a boy and a girl.
    Again, I am not sure about this. If I think of this in cases. Case 1: a girl is in the first spot (there are 4 choices) and a boy is in the last spot (there are 3 choices). There are 5 people left to disperse among the middle spots, 5!. To the first case we have 4*3*5!. Case 2: a boy in the first spot (3 choices) and a girl in the last spot (4 choices). There are 5 people left for the middle, 5!. So the answer to this case is also 3*4*5!. By the addition principle we have 4*3*5! + 3*4*5!, or we can write this as 2*3*4*5!, or we can write this as 4!5!, but this last way seems a bit harder to actually understand what you're counting.
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  6. #5  
    . DrRocket's Avatar
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    Quote Originally Posted by pbandjay
    Quote Originally Posted by Ots
    2. How many ways can they be arranged if a particular boy and a particular girl have to be next to each other (paired). I would see this as a pair, two boys and three girls or six 'things' so probably 6!
    I am not so sure about this. There is only a single way to actually arrange the boys/girls such that they are standing G B G B G B G. There are 4 spots for the girls, 4!. There are 3 spots for the guys, 3!. By the multiplication principle, the answer would be 4!3!. I just don't see how 6! limits the positions to boy girl boy girl...

    Quote Originally Posted by Ots
    3. If opposite sexes are at the ends. If the ends of G...B and B...G are considered to be the same, then 5! Basically, how many ways can five people be arranged in between a boy and a girl.
    Again, I am not sure about this. If I think of this in cases. Case 1: a girl is in the first spot (there are 4 choices) and a boy is in the last spot (there are 3 choices). There are 5 people left to disperse among the middle spots, 5!. To the first case we have 4*3*5!. Case 2: a boy in the first spot (3 choices) and a girl in the last spot (4 choices). There are 5 people left for the middle, 5!. So the answer to this case is also 3*4*5!. By the addition principle we have 4*3*5! + 3*4*5!, or we can write this as 2*3*4*5!, or we can write this as 4!5!, but this last way seems a bit harder to actually understand what you're counting.
    There is only problem with your analysis.

    You should be sure. That is because you are correct, and you have produced a valid (correct) argument to prove your position.
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  7. #6  
    Forum Ph.D. Heinsbergrelatz's Avatar
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    thank you to "pbandjay" and "ots"
    your analysis on the problem was very helpful, i appreciate it
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  8. #7  
    Ots
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    Pbandjay: you seem to want to alternate the boys and girls in your solution (BGBG...). But I don't see that as a requirement in the original problem.
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  9. #8  
    Forum Freshman pbandjay's Avatar
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    The way I understood the second restriction, they are alternating, which always guarantees there is not a boy not next to a girl (or a girl not next to a boy). I suppose I overlooked the situation. As the following situations all put a boy next to a girl:

    GBGBGBG
    BGGBGBG
    GBBGGBG
    etc...

    This makes the problem a bit more difficult to count. I still do not see how 6! will accurately count this, at first glance it looks like it would require the inclusion-exclusion principle at least. I propose to first count 7!, then subtract the amount of ways to group BBB together, then subtract the number of ways to group GGGG together, then add the number of ways to group BBBGGGG/GGGGBBB together, since these were subtracted out twice before. Then it is also possible to have a girl without a boy neighbor in a line up such as BGBGBGG and BGBGGGB, but not all of these types of cases were not subtracted out before.. some were such as in GGBBBGG. So this seems a bit more difficult than just 6!.

    I'm still trying to work on the final cases of this restriction.
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  10. #9  
    Ots
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    I took, "...a particular boy and a particular girl" to mean that one boy was paired with one girl. I see that you took it as boys and girls would all be paired.

    The question could have been clearer, but pairing leaves one out as there are seven to begin with. Also, just because they are paired doesn't mean they have to alternate, as you just noted.

    Anyway, if you pair one girl with one boy that leaves a pair and five other kids or six 'things.' How many ways can you order six things? I think it's 6! and that asssumes that each boy is unique and each girl is unique. If they aren't unique then it's a different formula.
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  11. #10  
    Forum Freshman pbandjay's Avatar
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    Okay, I am in agreement then. My last criticism is that yes there are 6 unique objects in a line, however 6! under counts this question. The "particular boy and girl" can either be lined up as BG or GB -- when they are counted as one object, only one of these cases is being counted. Therefore the 6! needs to be multiplied by two: 2*6!.

    Sorry for my confusion in this problem...
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  12. #11  
    Ots
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    Quote Originally Posted by pbandjay
    Okay, I am in agreement then. My last criticism is that yes there are 6 unique objects in a line, however 6! under counts this question. The "particular boy and girl" can either be lined up as BG or GB -- when they are counted as one object, only one of these cases is being counted. Therefore the 6! needs to be multiplied by two: 2*6!.

    Sorry for my confusion in this problem...
    Hey - no problem. I think the problem wasn't stated as clearly as it could have been.

    Yeah, 2*6! or 6! depending on whether GB paired is different than BG paired or not.
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