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Missy
science forum beginner
Joined: 02 Jul 2006
Posts: 5
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 Posted: Mon Jul 10, 2006 11:42 pm Post subject:
PLEASE explain this problem
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Like i have mentioned before I have not had math for some time now and
would like some assistance every now and again. would someone please
explain this problem to me.
4!/(2! × 2!) |
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David Moran
science forum Guru Wannabe
Joined: 13 May 2005
Posts: 252
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| Posted: Tue Jul 11, 2006 12:04 am Post subject:
Re: PLEASE explain this problem
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"Missy" <tawanda.green@yahoo.com> wrote in message
news:1152574969.832779.21250@35g2000cwc.googlegroups.com...
Like i have mentioned before I have not had math for some time now and
would like some assistance every now and again. would someone please
explain this problem to me.
4!/(2! × 2!)
Let n be some positive integer. Then n!=n(n-1)(n-2)(n-3)...(2)(1). So
4!=4(3)(2)(1)=24. Does this help?
Dave |
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Stratocaster
science forum addict
Joined: 13 Nov 2005
Posts: 63
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| Posted: Tue Jul 11, 2006 12:09 am Post subject:
Re: PLEASE explain this problem
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"Missy" <tawanda.green@yahoo.com> wrote in message
news:1152574969.832779.21250@35g2000cwc.googlegroups.com...
Like i have mentioned before I have not had math for some time now and
would like some assistance every now and again. would someone please
explain this problem to me.
4!/(2! × 2!)
{Yea! Finally something I can answer lol}
Ok, the Exclamation Mark represents a factorial. A factorial is the product
of all consecutive integers(whole numbers), starting from the integer
signified and goes down the line until 1.
For Example:
5! = 5*4*3*2*1 = 120
3! = 3*2*1 = 6
I'm sure you know this but just in case (I am using * above to indicate
multiplication) ; )
Just for the record
0! = 1
That is the way it is defined. Why would be for another thread.
Hopefully that is all you need to know. |
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jasen
science forum beginner
Joined: 28 Jun 2006
Posts: 16
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| Posted: Tue Jul 11, 2006 12:13 pm Post subject:
Re: PLEASE explain this problem
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On 2006-07-10, Missy <tawanda.green@yahoo.com> wrote:
| Quote: | Like i have mentioned before I have not had math for some time now and
would like some assistance every now and again. would someone please
explain this problem to me.
4!/(2! × 2!)
|
4! = 4 x 3 x 2 x 1
2! = 2 x 1
the answer is 6
--
Bye.
Jasen |
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matt271829-news@yahoo.co.
science forum Guru
Joined: 11 Sep 2005
Posts: 846
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| Posted: Tue Jul 11, 2006 1:06 pm Post subject:
Re: PLEASE explain this problem
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Missy wrote:
| Quote: | Like i have mentioned before I have not had math for some time now and
would like some assistance every now and again. would someone please
explain this problem to me.
4!/(2! × 2!)
|
This is not a "problem", it's an expression. Perhaps you meant to ask
what the notation means, in which case that's been answered. Otherwise,
what is the problem? |
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Darrell
science forum addict
Joined: 04 Jun 2005
Posts: 78
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| Posted: Wed Jul 12, 2006 5:37 am Post subject:
Re: PLEASE explain this problem
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"jasen" <jasen@free.net.nz> wrote in message
news:7bb9.44b39606.da153@gonzo.homenet...
| Quote: | On 2006-07-10, Missy <tawanda.green@yahoo.com> wrote:
Like i have mentioned before I have not had math for some time now and
would like some assistance every now and again. would someone please
explain this problem to me.
4!/(2! × 2!)
4! = 4 x 3 x 2 x 1
2! = 2 x 1
the answer is 6
|
I'll just add that sometimes (not necessarily this time, but there will be
times) it is useful to manipulate factorials such that they will cancel (vs.
expanding them out.)
Example: We could rewrite 4! as (4)(3)(2!) and we could further rewrite 4
as (2!)^2 giving
(3)(2!)^3 / (2!)^2
= (3)(2!)
= 6
--
Darrell |
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Missy
science forum beginner
Joined: 02 Jul 2006
Posts: 5
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| Posted: Wed Jul 12, 2006 10:49 pm Post subject:
thanks
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YES it helps.
David Moran wrote:
| Quote: | "Missy" <tawanda.green@yahoo.com> wrote in message
news:1152574969.832779.21250@35g2000cwc.googlegroups.com...
Like i have mentioned before I have not had math for some time now and
would like some assistance every now and again. would someone please
explain this problem to me.
4!/(2! × 2!)
Let n be some positive integer. Then n!=n(n-1)(n-2)(n-3)...(2)(1). So
4!=4(3)(2)(1)=24. Does this help?
Dave |
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