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The exercise says:

In the expansion $(2x+3y-4z+w)^9$, find the coefficient of $x^3y^2z^3$.

The formula to find the coefficient of $x_1^{r_1}x_2^{r^2}\dots x_k^{r_k}$ in $(x_1+x_2+\dots+x_k)^n$ is:

$$\frac{n!}{r_1!r_2!\dots r_k!}$$

Am I supposed to multiply the result determined by the above formula by $2$, $3$ and $4$? What to do when $x$ and $y$ has coefficients?

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The coefficient of $x^3y^2z^3$ is zero. Do you mean the coefficient of $x^3y^3z^3$? –  Michael Albanese Sep 23 '12 at 8:09
    
@MichaelAlbanese: Umm, no that's what I see in the book. What's wrong with zero? It's a nice number! –  Gigili Sep 23 '12 at 8:15
    
Fair enough, it just makes the question much easier. –  Michael Albanese Sep 23 '12 at 8:20
    
@MichaelAlbanese: Not everyone is as smart as you, and yes I Didn't notice it is zero and meant to ask the question for the general case. –  Gigili Sep 23 '12 at 8:26
1  
My comment wasn't intended to imply anything about my (or anyone else's) intelligence; if it came across as arrogant, I apologise but that was not my intention. I just thought that there may have been a typo (either yours or in the book). Given that you mentioned the multinomial coefficients, I assumed the exercise wanted you to use them, hence my question. –  Michael Albanese Sep 23 '12 at 9:02

2 Answers 2

up vote 4 down vote accepted

Every term in $(2x+3y-4z+w)^9$ is a product of nine factors, each of which is one of the four terms in parentheses. Thus, before you collect like terms each factor will have the form $(2x)^i(3y)^j(-4z)^kw^\ell$, where $i+j+k+\ell=9$. Since the exponents in $x^3y^2z^3$ add up to only $8$, not $9$, there is no such term in the product, and its coefficient is $0$.

If you actually meant the coefficient of $x^3y^3z^3$, each such term must arise as the product of three factors of $2x$, three of $3y$, and three of $-4z$, so it must be $(2x)^3(3y)^3(-4z)^3=2^33^3(-4)^3x^3y^3z^3$, with a coefficient of $2^33^3(-4)^3=-13824$. Your formulat tells you that there are

$$\frac{9!}{3!3!3!}=1680$$

such terms, so the total coefficient of $x^3y^3z^3$ is $-13824\cdot1680=-23~224~320$.

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Sum of exponents in $x^3 y^2 z^3$ equals $8$ :) –  M. Strochyk Sep 23 '12 at 8:17
    
@M.Strochyk: Tell my fingers to behave better! :-) (Thanks for catching it.) –  Brian M. Scott Sep 23 '12 at 8:18

Surely the question (although perhaps badly worded) is asking for the term in the expansion which contains $x^3y^2z^3$ as written? This term will be of the form $nx^3y^2z^3w$ for some integer $n$, so its "coefficient" is $nw$. We can expand

$(2x+3y-4z+w)^9 = ((2x+3y-4z)+w)^9$

as

$(2x+3y-4z)^9 + 9w(2x+3y-4z)^8 + \ldots$,

so $n$ is equal to 9 times the coefficient of $x^3y^2z^3$ in the expansion of $(2x+3y-4z)^8$. From Brian M. Scott's answer, we see that $n$ is equal to $$9 \times \frac{8!}{3!2!3!}$$ which I will let you calculate yourself. Also note that this is equal to $$\frac{9!}{3!2!3!1!}$$ which is the coefficient of $x^3y^2z^3w$, as expected.

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