SOLUTION: Show that: n C r = [(n-2) C r] + 2[(n-2) C (r-1)] + [(n-2) C (r-2)]

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Question 908843: Show that:
n C r = [(n-2) C r] + 2[(n-2) C (r-1)] + [(n-2) C (r-2)]

Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!


We will work with the right side:







Simplify and write letters before numbers:



Factor out (n-2)!



Write r! as r(r-1)(r-2)!
Write (r-1)! as (r-1)(r-2)!
Write (n-r)! as (n-r)(n-r-1)(n-r-2)!
Write (n-r-1)! as (n-r-1)(n-r-2)!



The LCD is 



(1)  

Now we simplify the numerator of (1):





We factor (n-r) out of the first two terms:







Distribute the (n-r)



Multiply everything out to remove all the parentheses:

{{n^2-r^2-n+r+r^2-r}}}

The r's and r²'s

}



Substitute for the numerator in (1)

(1)  

And when we multiply (n-2)! by that numerator,

(1)  

the numerator becones n!

(1)  

Next we simplify the LCD

LCD = 

That is just 

So now (1) becomes

(1)  

which is

(1)  

Whew!

Edwin

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