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Thread: can't figure these out..(word probs)

  1. #1

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    can't figure these out..(word probs)

    I got these at a math competition, and I have the answer to number one, but I have no clue how to do it. The answer is down below.

    1. How many ways can 6 men be seated in a row if two of the men must NOT be seated together?

    2. One of the lateral edges of a pyramid is 4 meters long. How far from the vertex will this edge be cut by a plane parallel to the base, which divides the pyramid into two equivalent parts (of equal volume). Round the final answer to the nearest .001 meter.






    ANSWER:

    1. 480
    - you've been privileged to read a post by Miz

  2. #2
    Fanatic Member prog_tom's Avatar
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    easy,

    6! - (6! / 3)

    or

    4*5!

    either way
    Last edited by prog_tom; Mar 8th, 2003 at 01:30 PM.

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  3. #3
    I don't do your homework! opus's Avatar
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    480???
    or as prog_tom said 6!/3 which is 240?????

    To place 6 man in a row

    First one 6 possibilities
    Second one 5
    Third one 4
    ....
    ...
    Last one 1

    Makes 6!=720 ways to place 6 man in a row!

    How many ways to place two man together in a row of six



    Second to 5th one can go either before or after that is 2 possibilities, except for the place 1 and 6 (no one before or after)

    So we have 4*2 +2*1 =10 ways to put man together in a row of six.

    So the answer should be 720 - 10 =710

    Or did I misunderstand something?
    ??????
    Last edited by opus; Mar 10th, 2003 at 09:40 AM.
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    I don't do your homework! opus's Avatar
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    prog_tom
    Re-editing a post after another one has answered is cheating.
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  5. #5
    I don't do your homework! opus's Avatar
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    And why should
    6! -(6!/3) and 2!5! be both the answer?
    The one is 480 the other is 240
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  6. #6
    Fanatic Member prog_tom's Avatar
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    I meant 4 * 5!, because there are combinations of 4 for these two persons(2^2) moving alone the line in the 5 people.

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  7. #7
    I don't do your homework! opus's Avatar
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    OK first of all I found the mistake in my approach.

    There are 6! poss. to place 6 people.

    There are 10 possibilities in which way you can place Person 1 and Person 2 sise-by side on those six places, BUT

    in each of these 10 cases, you have 4! poss. to place the other four.
    So it is 6! - (10*4!)=480
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    Addicted Member TheAlchemist's Avatar
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    hey guys,
    i don't think thats right logically,check thisout;
    the total number of ways we can arrange the 6 men with 2 of them being constantly together is 2*(5!)
    so the total number of ways in which they can be arranged where the same two men will not appear together is 6! -(2*(5!))
    which is equal to 480. same answer,differnent approaches!
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  9. #9
    I don't do your homework! opus's Avatar
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    Thanks TheAlchemist,
    I couldn't follow prog_toms way to calculate it, but now I get it!
    and I found the mistake in mine, coming to the same value, the problem is solved (for me at least)
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  10. #10
    Fanatic Member prog_tom's Avatar
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    Originally posted by TheAlchemist
    hey guys,
    i don't think thats right logically,check thisout;
    the total number of ways we can arrange the 6 men with 2 of them being constantly together is 2*(5!)
    so the total number of ways in which they can be arranged where the same two men will not appear together is 6! -(2*(5!))
    which is equal to 480. same answer,differnent approaches!
    Actually, it's the same approach man.

    123456 - 212345

    take out the common factors,

    12345(6-2)

    = 5!4

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  11. #11

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    sorry, i was outta town for the weekend so i couldn't reply, but thanks for the replies (the 6! - 2*5! makes sense the most to me). anyone know the second problem..or how to solve it?
    - you've been privileged to read a post by Miz

  12. #12
    Conquistador
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    read it wrong

    gimme a sec

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    Talking of word problems, you have to lol at my niece. I was helping her prepare for a math exam when she was still at school, and she was fighting with word problems. The one she was stuck on went something like "There are 2 towns a & b, 30 miles apart. Person A leaves a on his bike going towards b at 5 mph at 2pm, while person B leaves b for a at 2.15pm doing 6mph but goes via town c.... blah blah" (something like that, anyway, with more gumph thrown in for good measure.

    Finally she got to the actual question: "When did the latter get to town?"

    She threw her hands up in despair! "I can't do this! What's a 'latter'?"

    Well once I explained what the word meant, took her 2mins to do the sums. Half the time with these darned things, it's a problem with the English, not the Math!

  14. #14
    Fanatic Member prog_tom's Avatar
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    Originally posted by Spooner
    Half the time with these darned things, it's a problem with the English, not the Math!
    YEPPP!!!!!

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  15. #15

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    Originally posted by Spooner
    Half the time with these darned things, it's a problem with the English, not the Math!
    very true...i mean, why do people hate word probs anyway? not cuz of the math.
    - you've been privileged to read a post by Miz

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