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I am trying to understand a proof of counting number of trees using the conclusion of Cayley's formula.

At the end they got a $(x+y+5)^5$ and they say that the number of ways to get $xy^2$ from this is $5 \cdot \binom{4}{2} \cdot 5^2 = 750 $ and I can't figure how. Can someone please explain?

thanks.

4 Answers 4

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Here is a more basic explanation.

You have $5$ terms being multiplied, each of which is $(x+y+5)$.

To get $xy^2$ from the product, you have to choose which term to take $x$ from, and which two terms to take $y$ from. The remaining terms will contribute $5\times 5=25$.

There are $5$ possibilities to choose the term providing $x$. Once this is fixed, the two terms providing $y$ can be chosen in $\binom42 = 6$ ways. Hence we have the $xy^2$ term appearing $5\times6=30$ times, each time with a coefficient of $25$. Hence the coefficient of $xy^2$ in the expansion is $30\times25=750$.

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    What do you mean by: "The remaining terms will contribute $5×5 = 25$" ? – 2017-02-07
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    If you picked one $x$ and 2 $y$s, you have to pick the 5 from the other two terms. So the coefficient of each $xy^2$ picked is 25, and you have to multiply the 25 by the number of $xy^2$ to get the final coefficient of $xy^2$. – 2017-02-07
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    But If I have chosen one x from a term like $(x+y+5)$ and 2 y's from another two terms, this will give me $5 × 5 × 5$ from both two that I have mentioned. Or doesn't that count somehow? – 2017-02-07
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    From each term, you can choose an x or a y or a 5, not all 3. And you consider all possibilities of picking. So you end up with xxxxx+xxxxy+xxxyy+xxxy5+... and so on. What you are interested in is how many terms are integer multiples of $xy^2$, these look like rearrangements of xyy55. This would be the terms xyy55, xy5y5, 5yx5y and so on. My answer explains how to calculate their number and the total contribution. – 2017-02-07
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The coefficient of $y^2$ in $$((x+5)+y)^5$$ is $$\binom52(x+5)^3$$

In how many we shall find $x$ in $(x+5)^3?$

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The coefficient of $a^\alpha b^\beta c^\gamma$ in $(a+b+c)^n$ is:

$$\frac{n!}{\alpha!\beta!\gamma!}$$

where $\alpha+\beta+\gamma=n$.

We want $\alpha=1,\beta=2,\gamma=2$, which is $\dfrac{5!}{1!2!2!}=30$.

As $c=5$, we end with $5^2.30=750$.

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$$(x+y+z)^n \to \dfrac{n!}{k_1!k_2!k_3!}x^{k_1}y^{k_2}z^{k_3} ,k_1+k_2+k_3=n\\$$ $$(x+y+5)^5 \to \dfrac{5!}{k_1!k_2!k_3!}x^{k_1}y^{k_2}5^{k_3} ,k_1+k_2+k_3=5\\ \to xy^2 \to ,1+2+k_3=5 \to k_3=2\\ \dfrac{5!}{1!2!2!}x^{1}y^{2}5^{2} \\25\dfrac{5!}{1!2!2!}x^{1}y^{2}=25\times 30 xy^2$$