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What is the coefficient of the term $x^4 y^5$ in $(x+y+2)^{12}$?

How can we calculate this expression ?

I've applied the binomial theorem formula and got $91$ terms but I am not sure if it is right or wrong.

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  • $\begingroup$ This isn't a binomial, so you'll need something more than just a single use of the binomial theorem. $\endgroup$
    – vadim123
    Feb 28 '14 at 0:22
  • $\begingroup$ if its not a binomial then how can we calculate the expression ? $\endgroup$
    – Mora
    Feb 28 '14 at 0:25
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You could use the binomial theorem twice. Let $[x^{k}]$ denote the coefficient of $x^k$ from the polynomial $P(x)=\sum_{j=0}^{n}a_jx^j$, i.e. $[x^{k}]P(x)=a_{k}$. Now, $$\begin{eqnarray}[x^4y^5](x+y+2)^{12}&=&[x^4y^5](x+(y+2))^{12}\\&=&[x^4y^5]\sum_{j=0}^{12}\binom{12}{j}x^j(y+2)^{12-j}\\&=&[y^5]\binom{12}{4}(y+2)^{12-4}\\&=&\binom{12}{4}[y^5](y+2)^8\\&=&\binom{12}{4}[y^5]\sum_{j=0}^{8}\binom{8}{j}y^j2^{8-j}\\&=&\binom{12}{4}\binom{8}{5}2^{8-5}\\&=&\frac{12!}{4!8!}\frac{8!}{5!3!}2^3=\frac{12!2^3}{3!4!5!}\end{eqnarray}$$

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The trick is to apply trinomial formula here, the coefficient is $${12\choose4,5,3}2^3=\frac{12!2^3}{4!5!3!}$$

To see this, first consider $(x+y)$ as a whole body and then use binomial formula. Then the coefficient of term $(x+y)^9$ is $\displaystyle{12\choose 9}2^3=\frac{12!2^3}{9!3!}$ (Note we don't need to consider terms like $(x+y)^k$ other than $k=9$ here). And then consider the term $x^4y^5$ in $(x+y)^9$, whose coefficient shall be $\displaystyle{9\choose4}=\frac{9!}{4!5!}$. Combine two things, we obtain $${12\choose 9}2^3{9\choose4}=\frac{12!2^3}{9!3!}\frac{9!}{4!5!}=\frac{12!2^3}{4!5!3!}$$

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  • $\begingroup$ I haven't seen this formula before. Could you please explain how did you do it and got this coefficient ? $\endgroup$
    – Mora
    Feb 28 '14 at 0:33
  • $\begingroup$ @Mora See my edit plz. $\endgroup$
    – Shuchang
    Feb 28 '14 at 0:37
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For problems like this you can use The Multinomial Theorem: Let $n$ be a positive integer. For all $x_1,x_2,...,x_t$, $$(x_1+x_2+\cdots+x_t)=\sum {n\choose n_1,n_2, \cdots,n_t}x_{1}^{n_1}x_{2}^{n_2}\cdots x_t^{n_t},$$ where the summation extends over all nonnegative integral solutions $n_1,n_2,...,n_t$ of $n_1+n_2+\cdots n_t=n$.

So when $(x+y+2)^{12}$ is expanded, the coefficient of $x^4y^5$ is $${12\choose 4, 5, 3}(1)^4(1)^5(2)^3=221,760.$$

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