How can I find the Number of tagged trees $T=(V,E)$ with $V=\{ v_1, v_2, v_3, v_4, v_5, v_6, v_7, v_8, v_9, v_{10} \} $ such that $deg(v_i)=i$ for $1 \le i \le 3$ ?
- I need a way using Cayley's formula.
Thanks!
Number of tagged trees $T=(V,E)$ on $10$ vertices such that $deg(v_i)=i$ for $1 \le i \le 3$
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combinatorics
discrete-mathematics
graph-theory
1 Answers
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Just using Cayley's formula would be hard to find desired number of trees. But one of proofs of this formula should be useful. I mean proof based on analysis of Prüfer code. It is easy to see that $v$ appears in this code exactly $\deg v - 1$ times.
So all you need is just choose in the Prüfer code of your tree $0$ positions for $v_1$, $1$ position for $v_2$ and $2$ positions for $v_3$. All other positions can be occupied by any of $10 - 3 = 7$ remaining vertices. So the answer is $$\binom{8}{0, 1, 2, 8 - 0 - 1 - 2}\cdot (10 - 3)^{8 - 0 - 1 - 2} = \frac{8!}{0!\cdot 1! \cdot 2! \cdot 5!}\cdot 7^5 = 24 \cdot 7^6.$$
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0Can you explain why you multiply it by $(10−3)^{8−0−1−2}$ ? – 2017-02-07
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0It is number of possible ways to fill remaining $8 - 0 - 1 - 2$ positions in code with remaining $10 - 3$ vertices of unknown degrees. – 2017-02-07