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The coefficient of $x^{12}$ in $\left(x^{3}+x^{4}+x^{5}+x^{6}+\dots \right)^{3}$ is ___________.

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+30
\begin{align*} &\left [ {\color{red}{\bf x^{12}}} \right ] \left ( x^3 + x^4 + x^5 + x^6 + \dots \right )^3 \\ &\left [ {\color{red}{\bf x^{12}}} \right ] \left [ x^3\left ( 1 + x^1 + x^2 + x^3 + \dots \right ) \right ]^3 \\ &\left [ {\color{red}{\bf x^{12}}} \right ] \left [ x^9\left ( 1 + x^1 + x^2 + x^3 + \dots \right )^3 \right ] \\ &\left [ {\color{red}{\bf x^{3}}} \right ] \left [ \left ( 1 + x^1 + x^2 + x^3 + \dots \right )^3 \right ] \\ &\left [ {\color{red}{\bf x^{3}}} \right ] \left [ \left ( \frac{1}{1-x} \right )^3 \right ] \\ &\left [ {\color{red}{\bf x^{3}}} \right ] \left [ \sum_{k=0}^{\infty}\binom{3+k-1}{k}x^k \right ] \\ &\text{Now , put k = 3} \\ &\text{Coefficient of }\left [ {\color{red}{\bf x^{3}}} \right ] = \binom{2+3}{3} = 5C3 = 5C2 = 10 \\ \\ \hline \\ &\left [ {\color{red}{\bf x^{12}}} \right ] \left ( x^3 + x^4 + x^5 + x^6 + \dots \right )^3 \Rightarrow \bf 10 \\ \end{align*}
+24
This can be solved using common sense.

$(x ^3 + x^4 + x^5 + x^6 .....) ^ 3$

Terms beyond $x^6$  we cannot use to get x^12 here.

$(x ^3 + x^4 + x^5 + x^6 ) ^ 3$

We can use permutation

a,b,c -> take $x^a$ form term 1, $x^b$ form term 2 and $x^c$ form term 3

3,3,6 -> In 3 p3 / 2 ! ways = 3 ( as 3 is repreated so we have divided by 2!)

3, 4, 5 -> 3 p 3 ways = 6

4 , 4, 4 -> 3 p 3 / 3! ways = 1  ( 4 is repreated thrice so divided by 3!)

So total ways is  3 + 6 + 1= 10 ways.
+1
I am not able to solve generating function questions. How to solve them correctly?
0
+2

(x^3+x^4+x^5+x^6+…)^3 = 1/x^9 * (1+x+x^2+x^3+...)^3

now we have to find the coefficient of x^12/x^9 = x^3

this can be found in three ways - 3 OR 1+2 OR 1+1+1 = 3C1+3C1x2C1+3C3 =10

by Active
+1 vote

Another method:

When the given expression is expanded we will get 3 powers of x multiplying each other. We need to find out how many of these terms will give 12th power of x. That is we need to find the number of solutions for

x+y+z = 12-(3+3+3)

with the constrants: x>= 3, y>=3, z>=3 (since the powers of x start with 3 in the expression)

The problem reduces to find the number of solutions for the equation

x+y+z = 3

where x>=0, y>=0, z>=0

The number of solutions is 10 from the stars and bars method.

by Junior
We could answer this using generating functions or expanding the terms but they are time consuming.

Basically we want the number of ways to partition 12 into a sum of 3 integers, such that each integer must be 3 or greater. This is equivalent to writing $x_{1} + x_{2} + x_{3} = 12$ where $x_{i} \geq 3$. So that $x^{x_{1} + x_{2} + x_{3}} = x^{12}$ and each $x_{i}$ comes from one of the factors in in the cubic expansion.

So if I set $x_{i}' = x_{i} - 3$, we get $x_{1}' + x_{2}' + x_{3}' = 12 - 3.3 = 3$ and $x_{i}' \geq 0$. So this is the problem of distributing 3 identical balls in 3 distinct bins.

The answer becomes $C(3 + 3 - 1, 3) = C(5, 3) = 10$.
by Active
(x^3+x^4+x^5+x^6+- - -)^3

(x^3(1+x+x^2+x^3+- - -))^3

x^9(1+x+x^2+x^3+ - - -)^3

[note : (1+x+x^2+x^3+ - - -)

sum of infinite no: 1 / 1-r

r = x ]

==>x^9(1/(1-x)^3) = x^9((1-x)^-3)

==> x^9* 3-1+rCr            [ By n-1+rCr]

=> x^9*x^3(3-1+3C3)    [x^12 = x^9*x^3]

==> x^12(5C3)

==> 10x^12

Ans: 10