Show that there is a $6$ x $4$ board whose squares are all black or white, where no rectangle has the four vertex squares of the same color. Also show that on each $7$   x $4$ board whose squares are all black or white, there is always a rectangle whose four vertex squares are the same color. Point: On the board, adjacent houses do not necessarily have different colors.  I suspect it involves Pigeonhole principle or something discrete-mathematics, but I don't know how to solve it
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Suppose that each square of a $4 \times 7$ chessboard is colored either black or white. Prove that with any such coloring, the board must contain a rectangle (formed by the horizontal and vertical lines of the board) whose four distinct unit corner squares are all of the same color?   Any hints on this problem, I guess it could be solved with a clever application of the pigeonhole principle, but I have some difficulty seeing what the objects and what the bags are in which to put the objects to apply the pigeonhole principle, and then argue that such an coloring must exist. It is taken from a textbook on discrete mathematics, and in the text a version of the  is proven (using the pigeonhole principle). Hence, I guess it is somehow related to this.
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I was thinking if I could prove this relation by arranging the $(s − 1)(t − 1)$ in an $(s − 1) × (t − 1)$ 2-D grid and coloring an edge by color-1 if the two endpoints of the edge are in the same row and by color-2 if the two endpoints of the edge are in the same column. I'm not sure if this method will work, but any other approach would be really helpful.
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Can anyone recommend me an olympiad style combinatorics book which is suitable for a high schooler ? I know only some basics like Pigeon hole principle and stars and bars . I hope to find a  book which contains problems which purely test our originality ( the problems with beautiful constructions like , Nim problems, games,tillings, etc ) . More specifically problems which doesn't require theory but requires out of the box thinking . I don't know much about recurrence relations, generating functions or graph theory, so I would also love to see a book which introduces these topics .
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I wanted to share a really cool but simple problem.  Consider a coloring of the points of $\mathbb R^n$ with $n$ colors. Prove that there is a color $c$ such that for any $r&gt;0$ there are two points of the color $c$ at that distance.  Regards.
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I discovered  which claims that "$7$ is the only prime followed by a cube".  I find this statement rather surprising.  Is this true?  Where might I find a proof that shows this?  In my searching, I found , which is similar but the answers seem focused on squares next to cubes.  Any ideas?
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Prove:  If we color the plain with three different colours, then there will always be an equilateral triangle which has three vertices of the same colour.  I have proved it for two colours but I just can't solve this one.
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Prove that a perfect square is either a multiple of $4$ or of the form $4q+1$ for some $q\in \mathbb{Z}$.  Any ideas on how to start? Do I use a proof by contraposition? Also what's the definition of a perfect square?
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I'm having problems with this; the strategy is to show the left side is in the right side, and vice versa. Thus far I can only show that both sides of the equality are in A\B. Can I use "the empty set = A and not A" as a fact? Because then it works. Or just show me how to do it!
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I came across a question yesterday about combinations, and I wanted to know what the correct answer was. The question states as follows:  There are 8 spaces that are alternately black and white. There is one king, one queen, two identical rooks, two identical bishops, and two identical knights. The king needs to be surrounded by the two rooks. Then, the two bishops can be put on any of the remaining spaces as long as one is on a black space and the other is on a white space. Finally, the queen and the knights can be placed anywhere in the remaining spaces in relation to the other pieces. How many possible arrangements are there considering these conditions?
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On each cell of an $n \times n$ chessboard, we write a number from $1, 2, 3, .  . . , n$ in such a way that each number appears exactly $n$ times. Prove   that there exists a row or a column of the chessboard which contains   at least $\sqrt{n}$ distinct numbers.   This question is dying for a proof by contradiction, but I can't seem to show how. When I searched through the Internet, there is the following hint:     Assume each row and each column has less than $\sqrt{n}$ distinct numbers in   it. For each row or column, consider the number of distinct numbers in   it.   I feel like double counting is involved in the process, but what to count in two different ways? Also I can prove the statement with the probabilistic method but I would really like to learn the double counting approach.  Here is the other proof:  Choose a random row or column ($2n$ choices). Let $\textbf X$ be the number of distinct entries in it. Use the indicator variable $I_i$, for the existence of number $i$ in the block: $\textbf X = \sum I_i.$ Clearly, $E[I_i] = P[I_i = 1]$.  Then $P[I_i = 1] \geq 2\sqrt n / (2n) = 1/\sqrt n$. The lower bound is obtained if all the $i$ happens to be in some $\sqrt n$ edged square sub-matrix. Hence, by linearity $\textbf E[X] \geq \sqrt n$. This proves the existence.
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How many &quot;$1$&quot; can be selected at most, provided that only one from each row and column of the attached matrix $A$ is selected?(That is, when one &quot;$1$&quot; is selected, no other element can be selected than the row and column where that &quot;$1$&quot; is located.)  Answer: I think $6$ can be chosen.  How can I give this answer in connection with graph theory? I think it has to do with the Latin Square(or Rectangle). But I can't connect and explain. Please help me. I thought a lot, but I could not establish the relationship completely. Thanks...
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By contradiction. Suppose that any two vertices have an odd number of common neighbors. For an arbitrary vertex $v$, look at the subgraph induced by the neighbors of $v$. All degrees in this subgraph are odd(why all degrees are odd here?),   and hence the degree of $v$ is even(it is even because of the handshaking theorem?).   Let us count walks of length $2$ beginning at $v$. Their total number is even (because every vertex has an even degree). At the same time, an odd number of these walks come back to $v$, but any other vertex is reached by an odd number of walks, and hence the number of vertices distinct from $v$ must be even—a contradiction.  Please help me visualize this hint from my textbook.
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I'm bugged by the following that's summarized on p. 109 of .     False theorem: All horses are the same color.      Proof by induction:      $\fbox{$P(n)$ is the statement: In every set of horses of size $n$, all $n$ horses are the same color.}$   $\fbox{Base Case or $P(1)$:}$ One horse is the same color as itself. This is true by inspection.   $\fbox{Induction Step:}$ Assume $P(k)$ for some $k \geq 1$.   $\fbox{Proof of $P(k + 1) :$}$   Since $\{H_1, H_2, ... , H_n\}$ is a set of $n$ horses, the induction hypothesis applies to this set. Thus, all the horses in this set are the same color.   Since $\{H_2, H_3, ... , H_{n+1}\}$ is also a set of $n$ horses, the induction step likewise holds for this set. Thus, all the horses in this set are the same color too.         Therefore, all $n +1$ horses in $\{H_1, H_2, H_3, ... , H_n , H_{n+1}\}$ are the same color. QED.   The issue the instructor was trying to point out is clearly valid. For the case $n = 1$, there is only ${H_1}$. So this case says nothing about possible overlapping elements of each set of $(n + 1)$, for instance $H_2$ in the above proof.    But it was proposed in the class discussion that this was the only problem. Had you been able to prove $P(2)$ true, then a proof of the above format would have been fine.    My interpretation is that yes, you could prove all horses are the same color, if you can prove that any set of two horses will be the same color. But this format would not work.  Why not? The problem I see is that the above proof is for the existence of at least one particular pair of sets of horses of sizes $n$ and $n + 1$, such that in each set, all horses are the same color.  Particularly when the set of size $n$ is a subset of the set of size $n + 1$.  In order to prove the induction step, don't you need to prove that sets of sizes $n$ and $n + 1$, do not necessarily contain the same, overlapping elements?  You could prove that any horse can be added to a set of 2 horses.  Take the last two, and they must be the same color, and so on.  Wwhatever color the first two happen to be, all other horses must thus be the same color.    Am I misinterpreting the example, or am I making a logical error? Thanks in advance.
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I've seen two questions on solving the Rubik's cube but none of the answers have given a complete solution using mainly mathematical techniques. Furthermore, I've not seen a good explanation of general techniques for solving permutation puzzles in general, including those of the Rubik's cube family, without involving any memorized move sequences. So I've decided to write up my method below as an example of how group theory can be applied to solving such puzzles.
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A cube has numbers ±1 placed at each vertex and each center of cube’s six faces (so 14 numbers altogether). Moreover, it is known that each number in the center of a square face equals to the product of the 4 numbers at the vertices of this face. Can the sum of all 14 numbers be equal to zero? My first thoughts were test out some combinations and look for a pattern, but I was not able to notice anything major about it. So, the only progress I was able to really make was breaking it down into $V_1*V_2*V_3*V_4$, and so on for each face. Any help is greatly appreciated.
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In how many different ways can we place $8$ identical rooks on a chess board so that no two of them attack each other?  I tried to draw diagrams onto a $8\times8$ square but I'm only getting $16$ ways.  Does that sound right?  Thanks for the help!
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The question I am stuck on is 'how many ways can I choose $n$ balls from $3n$ balls, with $n$ indistinguishable black balls, $n$ indistinguishable white balls, and $n$ distinguishable coloured balls.'   I have been thinking about this problem for some time but can't get very far, I am used to questions about just distinguishable or indistinguishable objects, but I am not familiar with a problem involving both.  I think that there are $n+2$ 'colours' where within the colours all the balls are indistinguishable, but I don't really know where to go from there so any hints would be much appreciated.