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### Math Challenge II-A Combinatorics 6.29

Math Challenge II-A Combinatorics 6.29

Math Challenge II-A Combinatorics 6.29

We must also have 3k+l=73k+l=7,

Why 3k+L=7, though? Shouldn't it be 2k+L=7, for x^3 and x has respectively 3 & 1 power?

Thanks,

Andy

Re: Math Challenge II-A Combinatorics 6.29

I think $3\cdot k+1\cdot l=7$ is okay, as the powers of $x$ add up to $7$. We either have degree $3$ (from $x^3$) or degree $1$ (from $x$) and that's where the $3$ and $1$ come from here. $k$ represents how many degree $3$ and $l$ represents how many degree $1$ are used.

If that's still a little confusing, the main idea is that there are two ways to get $x^7$:

Case (i): $x^7$ comes from terms of the form $(x^3)^2\cdot x^1\cdot 2^2$ (we must have $2$ powers of $2$ so we have $5$ terms in total). This has a coefficient of $\dfrac{5!}{2!\cdot 1!\cdot 2!}$.

Case (ii): $x^7$ comes from terms of the form $(x^3)^1\cdot x^4\cdot 2^0$ (here there are no powers of $2$). This has a coefficient of $\dfrac{5!}{1!\cdot 4!\cdot 0!}$.

Hence, the final $x^7$ term is $$\frac{5!}{2!\cdot 1!\cdot 2!}\cdot (x^3)^2\cdot x^1\cdot 2^2 + \frac{5!}{1!\cdot 4!\cdot 0!}\cdot (x^3)^1\cdot x^4\cdot 2^0 = (120+5)x^7 = 125x^7.$$

Thanks a lot!!! I now got it

What can't k=0 and l=7 though?

Re: 回复： Re: Math Challenge II-A Combinatorics 6.29

Remember that our original expression is raised to the 5th power, so we can have at most 5 of any term.