SOLUTION: This is an intro to probability chapter in a book called Applied mathmatics. The book is a combo of two diferent books especially made for my school.
This is the problem:
Find
Algebra.Com
Question 149898: This is an intro to probability chapter in a book called Applied mathmatics. The book is a combo of two diferent books especially made for my school.
This is the problem:
Find (2n+2)!/(2n)!
I know how to do algebra, my problem I believe is because of the ! symbol. I have just been introduced to it and understand the concept , but not how it works in an agebraic expression.
Itried soving for n but don't know what to do with the !
ineed to understand the steps involved.
this is the answer:
4n2 +6n + 2
thanks in advance for any help.
heidi
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
The notation tells you to simply multiply that number by every integer that is less than that number until you hit 1. So if , this means that .
In this case, !=(2n+2)(2n+2-1)(2n+2-2)(2n+2-3)(2n+2-4)\ldots(3)(2)(1))
which simplifies to !=(2n+2)(2n+1)(2n)(2n-1)(2n-2)\ldots(3)(2)(1))
and !=(2n)(2n-1)(2n-2)(2n-3)\ldots(3)(2)(1))
So !}{(2n)!}=\frac{(2n+2)(2n+1)(2n)(2n-1)(2n-2)\ldots(3)(2)(1)}{(2n)(2n-1)(2n-2)(2n-3)\ldots(3)(2)(1)})
Highlight the common terms.
Cancel out the common terms.
!}{(2n)!}=(2n+2)(2n+1))
Simplify.
!}{(2n)!}=4n^2+6n+2)
FOIL to get
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