(New page: 2. What is the coefficient of x^3*y^6*z^5 in (x+y+z)^14? Explain in words why your answer is correct. I'm no good at memorizing arbitrary formulas, but I do know that this uses the multin...)
 
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14!/(3!*6!*5!) = 168168
 
14!/(3!*6!*5!) = 168168
  
[[Midterm Practice Problems]]
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[[Midterm Practice Questions]]

Revision as of 16:51, 9 March 2009

2. What is the coefficient of x^3*y^6*z^5 in (x+y+z)^14? Explain in words why your answer is correct.

I'm no good at memorizing arbitrary formulas, but I do know that this uses the multinomial theorem. Correct me if I'm wrong, but according to the theorem I think that the answer to this question would be

n!/(k_1)!*(k_2)!*(k_3)!

Where, in this example

n=14 k_1=3 k_2=6 k_3=5

So, the coefficient of x^3*y^6*z^5 in (x+y+z)^14 is

14!/(3!*6!*5!) = 168168

Midterm Practice Questions

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood