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#1 Re: Help Me ! » Polygon Diagonals permutation problem » 2007-08-17 06:21:38
Oh, ok. That's the logical explanation. I guess I got caught up with thinking that it was a combination/permutation type of problem.
That's interesting however. In order to discount duplicate diagonals, you divide. You divide by two because in this case, we're talking about lines that connect a vertex. In combination problems, division is often used to discount duplicates.
Thanks!
#2 Re: Help Me ! » Mixture problem » 2007-08-17 01:29:46
Bust out your Differential equations book!
I'd help, but i forget how to solve these kind of problems.
#3 Help Me ! » Polygon Diagonals permutation problem » 2007-08-17 01:18:37
- nsl22
- Replies: 2
Ok, I did a search on this and found this equation, but it doesn't explain anything.
The question is, How many sides does a Polygon with 20 diagonals have.
I've made drawings to figure it out, which obviously won't work for large polygons. So that's no good. I've also found a formula that says D = n(n-3)/2, where D is the number of diagonals and n is the number of sides. The polygon does indeed have 8 sides, but this is a permutation problem. Does anyone know of an analytical way of figuring this out?
Here's an easier question that is similar:
How many heptagons can be drawn by joining the vertices of a polygon with 10 sides?
A: heptagon = 7 sides. It's a Combination problem. You have 10 possible sides taken 7 at a time. So, 10*9*8*7*6*5*4/7! = 120
#4 Re: Help Me ! » License Plate type Permutation problem » 2007-08-17 00:57:14
That makes sense to me now. I read your simplified explanation and understood everything but where the 6 came from.
I figured it out though 6 = 4!/(2!*2!)
since there are 4 spaces and 2 #'s and 2 L's.
#5 Re: Help Me ! » A hard question that popped up in the exams 8years ago! » 2007-08-16 10:26:57
you could probably use this:
http://www.mathopenref.com/segmentarea.html
and this: http://www.mathopenref.com/arcsector.html
to come up with an answer.
#6 Re: Help Me ! » A hard question that popped up in the exams 8years ago! » 2007-08-16 10:05:41
I don't see the image "drawn as shown". I know what it looks like, but do not know what the shaded region is.
Nevermind.
What is: "(Pi - _/3)"???
#7 Re: Help Me ! » License Plate type Permutation problem » 2007-08-16 10:02:18
The answer is indeed 10000. There are 10000 distinct LL## permutations. If you allow the letters and numbers to combine in other orders (#L#L, ##LL, L#L#, etc), the total number of ways is even higher 60000 to be precise.
Can you explain? To me, it doesn't seem like you can have more than 10,000 combinations without getting duplicates.
#8 Help Me ! » License Plate type Permutation problem » 2007-08-16 08:58:22
- nsl22
- Replies: 8
Here is the problem that I cannot figure out 
. Please help me if you can.
Ten different letters of an alphabet are given. 2 of these letters followed by 2 digits are used to number the products of a company. In how many ways can the products be numbered?
The answer is 1000.
I'm confused by it because in order for it to be a valid combination, it has to be in the form of LL## where L = letter and # = number. If you say that for the first letter you have 10 choices, 2nd letter you ahve 10 choices, 1st number 10 choices and 2nd number 10 choices, you come up with 10*10*10*10 = 10,000 which is wrong because this is including the combinations of #L#L, ##LL, and L#L#.
How do you figure this out for only LL## combinations?
Thanks a lot!
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