Question 1159611
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Let *[tex \Large x] represent the number of 1st place finishes


Let *[tex \Large y] represent the number of 3re place finishes


Then the number of 2nd place finishes must be *[tex \Large x\,+\,y]


Given:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ (x\ +\ y)\ +\ y\ =\ 20]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5x\ +\ 3(x\ +\ y)\ +\ y\ =\ 68]


Simplify to a 2X2 system and solve (I did it by elimination, but any standard method will work).
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}
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