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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
easy,
6! - (6! / 3)
or
4*5!
either way
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?
??????
prog_tom
Re-editing a post after another one has answered is cheating.
:p :p :p :p
And why should
6! -(6!/3) and 2!5! be both the answer?
The one is 480 the other is 240
:confused: :confused: :confused:
I meant 4 * 5!, because there are combinations of 4 for these two persons(2^2) moving alone the line in the 5 people.
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
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!:( ;)
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)
Actually, it's the same approach man.Quote:
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!:( ;)
123456 - 212345
take out the common factors,
12345(6-2)
= 5!4
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?
read it wrong
gimme a sec :p
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!
YEPPP!!!!!:mad:Quote:
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.Quote:
Originally posted by Spooner
Half the time with these darned things, it's a problem with the English, not the Math!