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

+2 votes

(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
0 votes
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
0 votes
(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
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