# Tag Info

## Hot answers tagged combinatorics

22

For this kind of question, it's often best to make a table. Here's the data when put into a table. $$\begin{array}{c|cc|c} &\text{ Is Reasonable } & \text{ Not Reasonable } & \text{ Total} \\ \hline \text{Is Worse} &363 & ??? &487 \\ \text{Is Not Worse} & ??? & ??? & ???\\ \hline \text{Total} &756 & ??? & ... 5 Your proof is fine. An alternative for the factors of 3: we show that one of m,m+1,m+2 is divisible by 3 for any integer m, and then we can use that instead for your induction step for powers of 3. Basically, you need in your induction step to show that (4n+1)(4n+2)(4n+3)(4n+4) is divisible by both 8=2^3 and 3. It is divisible by 8 ... 3 You can always prevent one of them by colouring all edges the same color. Therefore it must be (inclusive) or. This gives, for instance, R(2,n)=n, since if there are n vertices, then either you have a K_2 of one color, or all edges are of the second color and you have K_n in that color. I think maybe the important point you've missed here is that ... 2 You're missing the requirement that the K_p or K_q must be completely red and blue, respectively. R(2,4) must be larger than 2 because if you color the single edge in K_2 blue, then the resulting graph does not contain a red K_2, and does not contain a blue K_4 either. 2 Your answer is correct. R\subset X\times X can be recognized as an equivalence relation. As comes forward in your question its equivalence classes are \{3,6,9\}, \{1,4,7,10\} and \{2,5,8\}. Then:$$R=\left[\{3,6,9\}\times\{3,6,9\}\right]\cup\left[\{1,4,7,10\}\times\{1,4,7,10\}\right]\cup\left[\{2,5,8\}\times\{2,5,8\}\right]$$And its cardinality ... 2 MIT has free lecture notes on combinatorics which are pretty concise and well written. I found it to be a good introduction (and I didn't need to attend the lectures to benefit from it). You can find it here. 2 It works. As an alternative, through a well-known formula:$$\nu_2((4n)!)=\sum_{m\geq 1}\left\lfloor\frac{4n}{2^m}\right\rfloor=\color{red}{3n}+\sum_{m\geq 3}\left\lfloor\frac{4n}{2^m}\right\rfloor\geq 4n-1-\log_2(n) $$and:$$\nu_3((4n)!)=\sum_{m\geq 1}\left\lfloor\frac{4n}{3^m}\right\rfloor\geq\color{red}{n}+\sum_{m\geq ...

2

I'll give you a few hints as to how you'd brute force solve this problem (there may be a smarter way to solve it): For any given $i$, we have $$a_i = \sum\limits_{j = 1}^i j(i - j) = \sum\limits_{j = 1}^i ji - j^2.$$ Use the formulae $\sum_{k = 1}^n k = \frac{n(n+1)}{2}$ and $\sum_{k = 1}^n k^2 = \frac{n(n+1)(2n + 1)}{6}$ to find a closed form for $a_i$. ...

1

Combinatorial Proof Let $N$ be the number of ways to select $n$ groups of four from $4n$ people. Then, if we label the groups by numbers $1$, $2$, $\ldots$, $n$, we can do it in $n!\cdot N$ ways. Alternatively, for $k=1,2,\ldots,n$, there are $4n-4(k-1)$ people from which four shall be chosen for the group with label $k$, and this can be done in ...

1

You may want to read Ivan Niven's Mathematics of Choice: How to Count Without Counting. It is a brief, accessible introduction to enumerative combinatorics that I would recommend as preliminary reading for students preparing to take an advanced undergraduate course in combinatorics. The text emphasizes problem solving and includes solutions to many of the ...

1

I found the article where this fact is shown. However, I can't access it in it's entirety without paying... Here it is It shows the first page, but everything else is kind of blurred out. It does not have the same layout as Ed Pegg's answer. I will try and edit this with a picture of the graph, but in the meantime, if maybe you have access to the above ...

1

Using the formula in my answer here, I calculate that $39480$ of the ${10\choose 2,2,2,2,2}=113400$ possible permutations have no consecutive letters the same.

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