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math challenge II-A Combinatorics Problem 6.9

math challenge II-A Combinatorics Problem 6.9

math challenge II-A Combinatorics Problem 6.9

The answer is (-1)^3 * 6 choose 3. I don't quite understand why do we multiple 6 choose 3 by -1 raised to the third power. If the constant term is y^3*(-2/y)^3, then is the answer (-2)^3*6 choose 3= -160?

Thanks!

Re: math challenge II-A Combinatorics Problem 6.9

You are correct that the (-1)^3 comes from the -1/y in the y - 1/y. It is true that the constant term of $(y - 2/y)^6$ is $-160$.