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May 25th, 2002, 10:54 PM
#1
Permutation + Combination
This is really easy for you. But not for me, cz I'm studying it after forever. 
Can someone explain to me permutation and combination?
I DO know that
nPk = n!/(n-k)!
nCk = n!(n-k)!k!
But I don't know where I'd be using these, as in what situations.
Any examples?
Thanks.
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May 25th, 2002, 11:05 PM
#2
Ok I found an example
PERMUTATION
Suppose that you walk into a theater that has a single row of 6 seats. If 2 of the seats are to be occupied, how many seating arrangements are possible?
The first person who walked into the theater had 6 seats to choose from and the second person had 5 choices. Answer is 6P2 = 30.
COMBINATION:
Consider the situation where the theater is so dark that the 2 individuals are indistinguishable. In this case the answer is 15, due to the fact that 2 different seating arrangements in the lighted theater account for a single arrangement in the dark theater.
Uhm... 
Perhaps I need another example, if anyone's got one, or two.
tnx.
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May 26th, 2002, 09:02 AM
#3
I'm surprised all you maths experts can't help with a concept as simple as this.
(fact: u r experts, I am dumb, so it's not so simple for me)
Anyone?
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May 26th, 2002, 03:07 PM
#4
In combinations the order is not important but in permutation it is. For example you have 6 books and you want to choose 3 books from them, you use combinations because it doesnt mather if you choose book 1-book2-book3, OR, book3-book1-book2 (doesnt mather what order you choose the books)
Another example: there are 10 christmas presents and you can choose 3 of them. How many different ways are there to choose 3 presents? again you dont care what order you choose the presents so you use combinations
Now lets say you want to find out how many ways are there to seat 4 people on a table, this time the order is important, so you use permutation..
sorry, my english is too bad to explain, I just used some examples
hth
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May 26th, 2002, 11:15 PM
#5
and my maths is poor 
OK, after the examples, I understood. Thanks.
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May 27th, 2002, 12:09 AM
#6
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May 29th, 2002, 06:43 AM
#7
Try this:
How about this:
You are at a Visual Basic Maths Forum, looking at the thread topics. You see three posts written by :
sql_lall (=s)
MrPolite (=M)
mendhak (=m)
How may ways are there to read two of the posts in order? (i.e. sql_lall then MrPolite, or MrPolite then sql_lall or mendhak then sql_lall)
= 3P2 = 3!/1! = 6
= Permutation (order important)
The 6 are:
s THEN M, M THEN s
s THEN m, m THEN s
m THEN M and M THEN m
How many ways are there to choose two authors and read their posts? (i.e. choose MrPolite and sql_lall, and read their posts in any order?)
= 3C2 = 3!/2!*1! = 6/2 = 3
= Combination (order not important)
The 3 are:
s AND M
s AND m
M AND m
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May 30th, 2002, 12:25 AM
#8
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May 31st, 2002, 06:38 AM
#9
OK...
Glad to help. Sorry, the last post was a bit corny, I know.
Well:
a)
The COMBINATIONS of picking all three
= the number of ways you can pick three different posts, and read them in any order.
= 3C3 = 3!/3!*1 = 6/6
= 1
i.e. pick s AND m AND M to read in any order.
b)
The PERMUTATIONS of picking all three
= the number of ways you can pick the three posts int the order you want to read them in (=pick the first post to read, then the next, then the last)
= 3P3 = 3!/0! -N.B. 0! = 1 (don't ask me why, it's just true)
= 6/1 = 6
These are:
1) s THEN m THEN M,
2) s THEN M THEN m,
3) m THEN s THEN M,
4) m THEN M THEN s,
5) M THEN s THEN m,
6) M THEN m THEN s
Notice that although all of these have s, m and M, that 1) differs from 2) becuase in 1), m is read before M, and in 2) M is read before m.
Hope this helps.
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May 31st, 2002, 11:48 AM
#10
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May 31st, 2002, 12:07 PM
#11
The factorial gives the number of ways in which n objects can be permuted. Since there is a single permutation of zero elements (the empty set called phi ), 0! = 1.
Factorials actually relate to permutations, not to how you calculate the arithmetic value of a factorial or the fact that 0 times anything is zero. It is not the same thing. Think of it as a definition.
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May 31st, 2002, 05:32 PM
#12
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Jun 2nd, 2002, 04:10 AM
#13
But wait, there's more...
If you thought 0! = 1 was weird, maybe you should sit down because:
(-0.5)! has a value. (The windows caluclator says it = root(Pi), but i don't think this is right)
=> Every positive integer AND every positive (Integer + 0.5) has a factorial.
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Jun 2nd, 2002, 05:20 AM
#14
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Jun 2nd, 2002, 08:24 AM
#15
Hyperactive Member
Look here at equations 6 - 11.
There are 10 types of people in the world - those that understand binary, and those that don't.
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Jun 2nd, 2002, 02:09 PM
#16
Originally posted by DavidHooper
Look here at equations 6 - 11.
tnx alot but I think I need to know more math before I learn that
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Jun 13th, 2002, 12:55 AM
#17
Re: Try this:
Originally posted by sql_lall
How about this:
You are at a Visual Basic Maths Forum, looking at the thread topics. You see three posts written by :
sql_lall (=s)
MrPolite (=M)
mendhak (=m)
How may ways are there to read two of the posts in order? (i.e. sql_lall then MrPolite, or MrPolite then sql_lall or mendhak then sql_lall)
= 3P2 = 3!/1! = 6
= Permutation (order important)
The 6 are:
s THEN M, M THEN s
s THEN m, m THEN s
m THEN M and M THEN m
How many ways are there to choose two authors and read their posts? (i.e. choose MrPolite and sql_lall, and read their posts in any order?)
= 3C2 = 3!/2!*1! = 6/2 = 3
= Combination (order not important)
The 3 are:
s AND M
s AND m
M AND m
That was a very good example. Thanks. I remember it like this: Combinations is just picking out a bunch, and Permutations is picking them out, but the order does matter.
thankz
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Jun 13th, 2002, 02:22 PM
#18
Hyperactive Member
I just remember permutations is the larger one.
There are 10 types of people in the world - those that understand binary, and those that don't.
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