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### Math Challenge II-A Combinatorics Chapter 6 Question 9

Math Challenge II-A Combinatorics Chapter 6 Question 9:

The answer is (-1)^3 * (6 choose 3). Why is it 6 choose 3 though? I couldn't quite understand the explanation from Mr. Lensmire in the video lecture. I know that the exponent of (y-1/y)^6 is 6, but why is it choose 3? And why is the coefficient of the constant term y^3 * (1/y)^3?

Thanks!

Let's take this a few steps at a time.

First, just to review the binomial theorem, which we are using here. It says that in $(A+B)^n$ the term containing $A^j B^{n-j}$ is $\displaystyle \binom{n}{j} A^j B^{n-j}$. As an example, $(A+B)^6$ contains the term $\displaystyle \binom{6}{3} A^3 B^3$. Does this part make sense?

Secondly, we need to apply the binomial theorem to the problem at hand. Note that $(y-1/y)^6$ is $(A+B)^6$ where $A = y$ and $B = - 1/y$. Hence (just like we had above), $(y-1/y)^6$ contains the term$$\binom{6}{3} A^3 B^3 = \binom{6}{3} (y)^3 (-1/y)^3 = - \binom{6}{3} = -20.$$

Lastly, why are we looking at the $(y)^3*(-1/y)^3$ term? Well, remember that all the terms are of the form $y^j (-1/y)^(6-j)$. How can this term have no $y$ variable (so it is a constant)? That only happens when the numerator and denominator have the same power of $y$, which is $y$.

I'm happy to explain more, just let me now if you have further questions!

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