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Mar 6th, 2003, 07:28 PM
#1
Thread Starter
Addicted Member
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
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Mar 7th, 2003, 12:49 AM
#2
Fanatic Member
easy,
6! - (6! / 3)
or
4*5!
either way
Last edited by prog_tom; Mar 8th, 2003 at 01:30 PM.

prog_tom
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Mar 7th, 2003, 03:01 AM
#3
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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Mar 7th, 2003, 03:06 PM
#4
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Mar 7th, 2003, 03:09 PM
#5
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Mar 8th, 2003, 01:32 PM
#6
Fanatic Member
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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Mar 10th, 2003, 09:39 AM
#7
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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Mar 10th, 2003, 10:07 AM
#8
Addicted Member
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!
One thing that sustains me through life is the conciousness of the immense inferiority of everyone else
--Oscar Wilde
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Mar 10th, 2003, 10:11 AM
#9
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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Mar 10th, 2003, 10:15 PM
#10
Fanatic Member
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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Mar 11th, 2003, 08:19 PM
#11
Thread Starter
Addicted Member
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?
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Mar 12th, 2003, 01:26 AM
#12
Conquistador
read it wrong
gimme a sec
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Mar 12th, 2003, 06:39 AM
#13
Addicted Member
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!
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Mar 13th, 2003, 09:18 PM
#14
Fanatic Member
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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Mar 14th, 2003, 05:21 PM
#15
Thread Starter
Addicted Member
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.
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