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8)a Code Is Formed It Consists Of 5 Characters
How Mny Different Codes Can Be Formed $*##$ .
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You have 5! because of the 5 characters (drawing a tree diagram will help explain), and dividing by 2!2! is because there are 2 repeated characters.
If n objects, comprising a of one type, b of another type, c of a third type, and so on, are arranged in a row, the total number of ways of doing it is given by
. If all the objects were different, there would be n! arrangements. Replacing a of these objects with an identical object means that a! (the number of ways of arranging the a objects replaced) arrangements are now identical. Thus we have counted each different arrangement a! times, so now we must divide by a!.Last edited by Toast (2006-12-20 21:48:25)
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Another way to explain is that you start out by choosing each character. There are two "$", so we choose 2 from 5:
C(5, 2)
Now there are 3 positions left. There is only 1 "*", so we choose 1 from 3:
C(3, 1)
Now there are 2 positions left. There are 2 "#", so we choose 2 from 2:
C(2, 2) = 1
Since each choice was independent, we may multiply these together to get the total. That is:
C(5, 2)*C(3, 1)*C(2, 2)
This turns out to be:
Which of course is the same answer as Toast.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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Very interesting guys! But I prefer to iterate all the ways until I become more familiar with it.
igloo myrtilles fourmis
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That's a perfectly good way of doing it as long as your method of finding all of the iterations is sound, and probably the easiest way if you're not sure about all the factorial formulas.
However, due to the nature of factorials, the numbers of combinations will get very high very quickly and so for certain situations, your way would take a very long time.
Why did the vector cross the road?
It wanted to be normal.
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I actually disagree. What do you do if your problem is incredibly huge? Say something in the millions of things? Then you must come up with a way to count, instead of doing an exhaustive pattern.
Doing it on smaller problems is for the purpose of teaching how to count combinations. Exhaustive searching ignores this lesson.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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I prefer to call "exhaustive" searching "fun and enlightening" searching.
igloo myrtilles fourmis
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Ah, by exhaustive I just meant that you find all of the possibilities.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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OH yeah, that is an actual term, but it makes it sound like you get exhausted, which you could if you weren't trying different approaches and various groupings that you haven't tried before.
If you want to see some exhaustive work, check out the diagram I made by hand at:
johnericfranklin.250free.com
See all those grids filled in, there is rhyme and reason to it all. And I thought it was a "blast!!!"
Last edited by John E. Franklin (2006-12-21 16:57:55)
igloo myrtilles fourmis
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