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Missy
science forum beginner

Joined: 02 Jul 2006
Posts: 5

Posted: Wed Jul 12, 2006 10:49 pm    Post subject: thanks

YES it helps.

David Moran wrote:
 Quote: "Missy" 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
Darrell

Joined: 04 Jun 2005
Posts: 78

Posted: Wed Jul 12, 2006 5:37 am    Post subject: Re: PLEASE explain this problem

"jasen" <jasen@free.net.nz> wrote in message
news:7bb9.44b39606.da153@gonzo.homenet...
 Quote: On 2006-07-10, Missy 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
matt271829-news@yahoo.co.
science forum Guru

Joined: 11 Sep 2005
Posts: 846

Posted: Tue Jul 11, 2006 1:06 pm    Post subject: Re: PLEASE explain this problem

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?
jasen
science forum beginner

Joined: 28 Jun 2006
Posts: 16

Posted: Tue Jul 11, 2006 12:13 pm    Post subject: Re: PLEASE explain this problem

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
Stratocaster

Joined: 13 Nov 2005
Posts: 63

 Posted: Tue Jul 11, 2006 12:09 am    Post subject: Re: PLEASE explain this problem "Missy" 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.
David Moran
science forum Guru Wannabe

Joined: 13 May 2005
Posts: 252

 Posted: Tue Jul 11, 2006 12:04 am    Post subject: Re: PLEASE explain this problem "Missy" 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
Missy
science forum beginner

Joined: 02 Jul 2006
Posts: 5

 Posted: Mon Jul 10, 2006 11:42 pm    Post subject: PLEASE explain this problem 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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