Difference between revisions of "Multinomial Theorem"

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<math>\binom{n}{k_1,k_2,\cdots,k_x}=\dfrac{n!}{k_1!k_2!\cdots k_x!}</math>
 
<math>\binom{n}{k_1,k_2,\cdots,k_x}=\dfrac{n!}{k_1!k_2!\cdots k_x!}</math>
  
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==Problems==
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===Introductory===
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===Intermediate===
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*The expression
  
== Intermediate Problems ==
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<math>(x+y+z)^{2006}+(x-y-z)^{2006}</math>
  
* [[2006_AMC_12A_Problems/Problem_24 | 2006 AMC 12A Problem 24]]
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is simplified by expanding it and combining like terms. How many terms are in the simplified expression?
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<math> \mathrm{(A) \ } 6018\qquad \mathrm{(B) \ } 671,676\qquad \mathrm{(C) \ } 1,007,514</math><math>\mathrm{(D) \ } 1,008,016\qquad\mathrm{(E) \ }  2,015,028</math>
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-([[2006_AMC_12A_Problems/Problem_24|2006 AMC 12A Problem 24]])
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===Olympiad===
  
  
 
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Revision as of 20:54, 24 September 2007

The multinomial theorem states that

$(a_1+a_2+\cdots+a_x)^n=\sum_{k_1,k_2,\cdots,k_x}\binom{n}{k_1,k_2,\cdots,k_x}a_1^{k_1}a_2^{k_2}\cdots a_x^{k_x}$

where

$\binom{n}{k_1,k_2,\cdots,k_x}=\dfrac{n!}{k_1!k_2!\cdots k_x!}$

Problems

Introductory

Intermediate

  • The expression

$(x+y+z)^{2006}+(x-y-z)^{2006}$

is simplified by expanding it and combining like terms. How many terms are in the simplified expression?

$\mathrm{(A) \ } 6018\qquad \mathrm{(B) \ } 671,676\qquad \mathrm{(C) \ } 1,007,514$$\mathrm{(D) \ } 1,008,016\qquad\mathrm{(E) \ }  2,015,028$

-(2006 AMC 12A Problem 24)

Olympiad

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