What is the coefficient of x^4y^4 in the expansion of (x+y)^8

Algebra Examples

Expand Using the Binomial Theorem (x+4y)^4

Step 1

Use the binomial expansion theorem to find each term. The binomial theorem states .

Step 3

Simplify the exponents for each term of the expansion.

Step 4

Apply the product rule to .

Rewrite using the commutative property of multiplication.

Rewrite using the commutative property of multiplication.

Apply the product rule to .

Rewrite using the commutative property of multiplication.

Apply the product rule to .

Rewrite using the commutative property of multiplication.

Multiply by by adding the exponents.

Use the power rule to combine exponents.

Apply the product rule to .

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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.

asked Feb 28, 2014 at 0:18

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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, $$\begin{eqnarray}[x^4y^5](x+y+2)^{12}&=&[x^4y^5](x+(y+2))^{12}\\&=&[x^4y^5]\sum_{j=0}^{12}\binom{12}{j}x^j(y+2)^{12-j}\\&=&[y^5]\binom{12}{4}(y+2)^{12-4}\\&=&\binom{12}{4}[y^5](y+2)^8\\&=&\binom{12}{4}[y^5]\sum_{j=0}^{8}\binom{8}{j}y^j2^{8-j}\\&=&\binom{12}{4}\binom{8}{5}2^{8-5}\\&=&\frac{12!}{4!8!}\frac{8!}{5!3!}2^3=\frac{12!2^3}{3!4!5!}\end{eqnarray}$$

answered Feb 28, 2014 at 21:57

epi163sqrtepi163sqrt

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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 $${12\choose 9}2^3{9\choose4}=\frac{12!2^3}{9!3!}\frac{9!}{4!5!}=\frac{12!2^3}{4!5!3!}$$

answered Feb 28, 2014 at 0:29

ShuchangShuchang

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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 $${12\choose 4, 5, 3}(1)^4(1)^5(2)^3=221,760.$$

answered Feb 28, 2014 at 17:02

1233dfv1233dfv

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Coefficient of#x^8 y^5# is#1287#

Explanation:

We know #(a+b)^n= nC_0 a^n*b^0 +nC_1 a^(n-1)*b^1 + nC_2 a^(n-2)*b^2+..........+nC_n a^(n-n)*b^n#

Here #a=x,b=y,n=13# We know, #n C_r = (n!)/(r!*(n-r)!#

#x^8 *y^5# will be in #6# th term

#T_6= 13 C_5 * x^(13-5)* y^5= 13 C_5 x^8*y^5#

#13 C_5 =(13!)/(5!*(13-5)!) =1287#

Coefficient of #x^8 y^5# is #1287# [Ans]

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