Binomial Expansion: Coeff of x^29 in (1+x^5+x^7+x^9)

[icystrike]

Homework Statement

Find the coefficient of $$x^{29}$$ in the expansion of $$(1+x^{5}+x^{7}+x^{9})$$.

Homework Equations

The Attempt at a Solution

 

[CompuChip]

It is zero: there is no $x^{29}$ term in
$$(1+x^{5}+x^{7}+x^{9})$$

I suppose you are missing some power?

 

[icystrike]

It is zero: there is no $x^{29}$ term in
$$(1+x^{5}+x^{7}+x^{9})$$

I suppose you are missing some power?

oh yea sorry.
its
$$(1+x^{5}+x^{7}+x^{9})^{16}$$

 

[CompuChip]

So you have 16 factors of
$$(1+x^{5}+x^{7}+x^{9})$$
multiplying. The first thing I'd do is check which unique combinations of powers give 29. For example, suppose that you open up the brackets. You will encounter terms with four factors of x^5, a factor of x^9 and all 1's otherwise, which gives x^29. Are there any other combinations?

Then, go through them one by one... you are looking for something of the form
$$x^9 \cdot x^5 \cdot x^5 \cdot x^5 \cdot x^5 \cdot 1 \cdot 1 \ cdot 1 \cdots$$
(16 in total). How many different orders are there? I.e., when you again think about multiplying out the brackets, how many terms are there that give this expression?

 

[icystrike]

I tried using another method:

 

[CompuChip]

Yeah, well, that was essentially what I was thinking of too.

And your answer is correct.

 

[icystrike]

thanks compuchip (=