Multinomial Expansion: Coefficient & Term Count for (w+x+y+z)^{23}

[Mentallic]

Homework Statement

1) Find the coefficient of w^2x^5y^7z^9 in (w+x+y+z)^{23}
2) When (w+x+y+z)^{23} is expanded and terms collected, how many different terms will there be?

The Attempt at a Solution

Again, I'm studying through my textbook and I'm finding these kinds of questions in past exams that aren't present in the book. I'm unsure how to begin.

 

[micromass]

Hi mentallic!

Do you know the multinomial theorem? This answers your question immediately.

See http://en.wikipedia.org/wiki/Multinomial_theorem

Basically, it states that

(x_1+...+x_m)=\sum_{k_1+...+k_m=n}{\binom{n}{k_1,...,k_m}x_1^{k_1}...x_m^{k_m}}

So what does that give us in our specific case? Here: m=4 and n=23.

 

[Mentallic]

Oh thanks, that makes finding the answer very simple!

But the multinomial expansion isn't in our syllabus so I'm guessing we need to argue with separate combinatoric multiplications.

Judging by the multinomial expansion though, I'm guessing the second last step in the solution would be of the form:

\binom{23}{2}\binom{23-2}{5}\binom{23-2-5}{7}\binom{23-2-5-7}{9}=\frac{23!}{2!5!7!9!}

Somehow I need to be able to argue why the coefficient is equal to these combinations?

And I still have no clue how to answer part 2.

 

[micromass]

OK, the things that you could do then is actually show the multinomial theorem in the case m=4. This is very easy and natural because it just requires you to use the binomial theorem a few times.

So, let's start with

(y+z)^{23}

this is just the binomial theorem. Then we add one more term:

(x+(y+z))^{23}

this is again using the binomial theorem and previous result. And finally, we have

(w+(x+y+z))^{23}

Working this out should give you the statement of the multinomial theorem.

Now, this statement is also needed in part 2, since that basically asks you to calculate in how many ways we can write k_1+k_2+k_3+k_4=n.

Oh thanks, that makes finding the answer very simple!

But the multinomial expansion isn't in our syllabus so I'm guessing we need to argue with separate combinatoric multiplications.

Judging by the multinomial expansion though, I'm guessing the second last step in the solution would be of the form:

\binom{23}{2}\binom{23-2}{5}\binom{23-2-5}{7}\binom{23-2-5-7}{9}=\frac{23!}{2!5!7!9!}

Somehow I need to be able to argue why the coefficient is equal to these combinations?

And I still have no clue how to answer part 2.

 

[Mentallic]

OK, the things that you could do then is actually show the multinomial theorem in the case m=4. This is very easy and natural because it just requires you to use the binomial theorem a few times.

So, let's start with

(y+z)^{23}

this is just the binomial theorem. Then we add one more term:

(x+(y+z))^{23}

this is again using the binomial theorem and previous result. And finally, we have

(w+(x+y+z))^{23}

Working this out should give you the statement of the multinomial theorem.

Hmm, since we haven't diverged from the multinomial theorem, I'll use it, but I don't know how happy they'll be about me using a different technique to the one they taught in the class that I missed.
It'll suffice though. I mean, an answer's an answer, right?

Now, this statement is also needed in part 2, since that basically asks you to calculate in how many ways we can write k_1+k_2+k_3+k_4=n.

Oh really? Ok well that I can answer, but I don't quite understand why exactly. I'll have to come back to this detail after my exam is over.

Thanks a lot for your help micromass!

 

[NguyenSP]

When (w+x+y+z)²³ is expanded and terms collected, how many different terms will there be?

from pascal's triangle you can find \frac{26!}{23!3!}= 5200 diffenent terms.