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Question 8.13 MCIIA Combinatorics

 
 
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Re: Question 8.13 MCIIA Combinatorics
by David Reynoso - Wednesday, 8 April 2020, 6:23 PM
 

For these kinds of questions we want to use the multinomial theorem.

For three variables the multinomial theorem looks like this: the coefficient of $a^d\cdot b^e\cdot c^f$, with $d + e + f = n$, in $(a + b + c)^n$ is $$\dfrac{n!}{d! \cdot e! \cdot f!}.$$

So, for example, the coefficient of $x^2 y^3 z$ in $(x + y + z)^6$ would be $$\dfrac{6!}{2!\cdot 3! \cdot 1!} = 60,$$ and the coefficient of $x y^4 z$ would be $$\dfrac{6!}{1! \cdot 4! \cdot 1!} = 30.$$