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

3 votes
3 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

3 votes
3 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 $C(3 + 3 - 1, 3) = C(5, 3) = 10$.
3 votes
3 votes
This problem is same as solving the below problem

Find total number of possible solution for given equation:

$x_1 + x_2+ x_2 = 12$

$x_i \geq 3, for\ i=1,2,3$

(Can be seen as number of ways to distribute 12 indistinguishable balls into 3 distinguishable bins, where each bin gets atleast 3 balls)

 

lets first put 3 in every box,

In total 3*3 = 9 numbers are gone

 

so above problem then further can be converted into:

number of ways to distribute 3 indistinguishable balls into 3 distinguishable bins, where each bin gets zero or more balls

$x_1 + x_2+ x_2 = 3$

$x_i \geq 0, for\ i=1,2,3$

soltuion to this problem is straight forward:

$\binom{3+3 -1}{3} = \binom{5}{3} = \binom{5}{2} = 10$ ways.
2 votes
2 votes

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.

Answer:

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