+0  
 
+5
402
6
avatar+238 

5C2=?   

so the formlar is nCr=n!/[r!*(n-r)]!,right?So,5C2=5!/[2!(5-2)]!=5*4*3*2*1/(2*1*3*2*1)=120/12=10

 

But I can do it in another way 5*4*/(1*2)=10

 

same other problem,such as 7C5=7!/(5!*(7-5)!)=21

 

7*6*5*4*3/(5*4*3*2*1)=21

 

so i can write my method in

 

nCr=n(n-1)(n-2)(n-3)......(n-r)/r*(r-1)(r-2)(r-3).....2*1

 

can someone prove my method whether true or false?if true , why it equal to the original formula? 

quinn  Oct 9, 2014

Best Answer 

 #4
avatar+26718 
+10

There is a function called the Gamma function, defined as follows:

 

                                             $$\Gamma (t)=\int_{0}^{\infty}x^{t-1}e^{-x}dx$$

 

When t is a positive integer this is equal to (n - 1)! but the Gamma function is defined for all values of t, so it is sometimes thought of as the extended factorial function.  

 

So 5.6! would be calculated from Γ(6.6) (≈ 344.702...)   (5! = 120,  6! = 720)

 

.

Alan  Oct 9, 2014
 #1
avatar+17736 
+5

I see two parts to you question:

1)  If you want 2 men and 5 women, you wouldn't use 5C2;  5C2 represents the number of ways that you can        choose a set of 2 out of a set of 5;  this does not represent choosing 2 men and 5 women.

2)  Your formula:

     nCr  =  n! / [ r! · (n-r)! ]

     The  (n-r)!  in the denominator cancels all the factors in the numerator starting with (n-r) down through 1;        leaving only  n · (n-1) · (n-2) · ... · (n-r+2) · (n-r+1).

     There is still a factor of  r!  in the denominator.

     I agree with your process EXCEPT I feel that you need one fewer term in the numerator, because the (n-r)        factor gets cancelled also; in the numerator, stop with factor (n-r+1).

geno3141  Oct 9, 2014
 #2
avatar+238 
0

okay the original question is In a trilsupose a lawyerwants 2 men and 5 womento make up a special panel.if the 7 panel  menbers are selected at random fro a pool of 12 peoples,which istheprobality that the  lawyer will not get the desired panel?(sorry for i didnt write the question completelyand i got the anwser)

ok i got it know.Thank you for your help.

why 0!=1? can n be a negative or fraction in n! ?why or why not?

quinn  Oct 9, 2014
 #3
avatar+92623 
+10

Hi Quinn,

I have heard of an extended definition of factorial that does allow this but I have never used it.

From my knowledge base factorials belong to whole positive integers and 0 only.

5! means 1*2*3*4*5

What would 5.6! mean?    It means nothing to me.  

Melody  Oct 9, 2014
 #4
avatar+26718 
+10
Best Answer

There is a function called the Gamma function, defined as follows:

 

                                             $$\Gamma (t)=\int_{0}^{\infty}x^{t-1}e^{-x}dx$$

 

When t is a positive integer this is equal to (n - 1)! but the Gamma function is defined for all values of t, so it is sometimes thought of as the extended factorial function.  

 

So 5.6! would be calculated from Γ(6.6) (≈ 344.702...)   (5! = 120,  6! = 720)

 

.

Alan  Oct 9, 2014
 #5
avatar+92623 
0

Thanks Alan,  I knew this existed but I have never used it.  :)

http://www.wolframalpha.com/input/?i=gamma%20function

Melody  Oct 9, 2014
 #6
avatar+238 
0

okay! Thank you,guys.

quinn  Oct 10, 2014

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