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Help me in this simple prob..


Prashan_punk
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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]

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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 by alan2here
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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.

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