CAN ANY ONE PROVE THIS?

 

 

( n! ) /( r! (n-r)! ) = ( n(n-1) )/ 2!

Edited September 27, 200817 yr by Prashan_punk

It doesn't hold. For example, let n = 10 and r = 3.

No, but I can find plenty of counterexamples.

sorry guys about some mistake ....but i am still confuse in this binomial expansion.....can u explain that part =>HOW IS C(n,2)x^2 = (n(n-1))/2!) OR C(n,3)x^3 = (n(n-1)(n-2))/(3!) of this equation....

See http://en.wikipedia.org/wiki/Binomial_theorem#Proof for a proof (by induction) of the binomial theorem. This appears only to prove the case where n is a positive integer. To be honest, I wouldn't mind seeing a proof of the general (real or complex) case myself.

It just follows from the definition of nCr. For example, for nC3:

 

[math]\binom{n}{k} = \frac{n!}{k!(n-k)!} \Rightarrow \binom{n}{3} = \frac{n!}{3!(n-3)!}[/math]

 

Now if we write this out, simply notice that most of the terms on the top and bottom cancel each other out:

 

[math]\binom{n}{3} = \frac{1}{3!} \cdot \frac{n \cdot(n-1) \cdot (n-2) \cdot (n-3) \cdot (n-4) \cdots 2 \cdot 1}{(n-3) \cdot (n-4) \cdots 2 \cdot 1} = \frac{n(n-1)(n-2)}{3!}[/math]

What do the !'s mean?

Factorial.

TY Refsmmat :¬)

 

I understand

(n!)/(r!(n-r)!) = (n(n-1)) / 2!

 

To mean

n!/(r!*((n-r)!)) = (n*(n-1)) / 2

 

and that I could rearrange it like this

n!/(r!*((n-r)!)) - ((n*(n-1)) / 2) = 0

 

If n = 1 and r = 1

n!/(r!*((n-r)!)) = (n*(n-1)) / 2

1/0 = 0/2

inf = 2

 

If n = r

n!/(r!*((n-r)!)) = (n*(n-1)) / 2

n!/0 = (n*(n-1)) / 2

inf = (n*(n-1)) / 2

 

if n = 4 and r = 2

n!/(r!*((n-r)!)) = (n*(n-1)) / 2

(4*3*2)/(2*(((4*3*2)-2)!)) = ((4*3*2)*((4*3*2)-1)) / 2

24/(2*(22!)) = (24*23) / 2

<1 = 276

 

Have I got all my maths correctly and does this prove the equation incorrect?

Edited October 7, 200817 yr by alan2here

0! is defined to be equal to 1, not 0. (Didn't check the rest, but I noticed that error straight off.)

Thanks. Let me try again with the last part of my post.

 

If n = 1 and r = 1

n!/(r!*((n-r)!)) = (n*(n-1)) / 2

1 = 0

 

If n = r

n!/(r!*((n-r)!)) = (n*(n-1)) / 2

n!/(r!*1) = (n*(n-1)) / 2

 

if n = 4 and r = 2

n!/(r!*((n-r)!)) = (n*(n-1)) / 2

(4*3*2)/(2*2) = (4*3) / 2

24/4 = 12/2

6 = 6

 

So possibly

n!/(r!*((n-r)!)) = (n*(n-1)) / 2

Is true in some cases.

nice question and nice answer by Alen