Prashan_punk Posted September 27, 2008 Share Posted September 27, 2008 (edited) CAN ANY ONE PROVE THIS? ( n! ) /( r! (n-r)! ) = ( n(n-1) )/ 2! Edited September 27, 2008 by Prashan_punk Link to comment Share on other sites More sharing options...
Josy Posted September 27, 2008 Share Posted September 27, 2008 It doesn't hold. For example, let n = 10 and r = 3. Link to comment Share on other sites More sharing options...
the tree Posted September 27, 2008 Share Posted September 27, 2008 No, but I can find plenty of counterexamples. Link to comment Share on other sites More sharing options...
Prashan_punk Posted September 28, 2008 Author Share Posted September 28, 2008 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.... Link to comment Share on other sites More sharing options...
Josy Posted September 28, 2008 Share Posted September 28, 2008 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. Link to comment Share on other sites More sharing options...
Dave Posted September 29, 2008 Share Posted September 29, 2008 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] Link to comment Share on other sites More sharing options...
alan2here Posted October 6, 2008 Share Posted October 6, 2008 What do the !'s mean? Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted October 6, 2008 Share Posted October 6, 2008 Factorial. Link to comment Share on other sites More sharing options...
alan2here Posted October 7, 2008 Share Posted October 7, 2008 (edited) 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, 2008 by alan2here Link to comment Share on other sites More sharing options...
Bignose Posted October 7, 2008 Share Posted October 7, 2008 0! is defined to be equal to 1, not 0. (Didn't check the rest, but I noticed that error straight off.) Link to comment Share on other sites More sharing options...
alan2here Posted October 8, 2008 Share Posted October 8, 2008 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. Link to comment Share on other sites More sharing options...
zouhib Posted October 8, 2008 Share Posted October 8, 2008 nice question and nice answer by Alen Link to comment Share on other sites More sharing options...
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