We know that the ways to go from A to B in a $m×n$ rectangle is $\frac{(m+n)!}{m!n!}$.  then find ways to go from $A$ to $B$ just by using this formula.(shortest way)     My Attempt:I calculated the $4*4$ square ways there are 70 ways now we should reduce the ways that should'nt calculated.Can you help me please??
121884
We know that the ways to go from A to B in a $m×n$ rectangle is $(m+n)!/m!n!$ I want a proof for this formula.(the way should be the shortest.  My Attempt:There is a counting way that may help us.   Every point is calculated by the sum of two adjacent points.
5523
(a)For $m,n \in \mathbb{N}$, $m&lt;n$ and $k=0, ..., m$: $$\frac{1}{m^k} \binom{m}{k} \leq \frac{1}{n^k} \binom{n}{k}$$ (b)For $n\in \mathbb{N}$ and $k=1, ..., n:$ $$\frac{1}{n^k}  \binom{n}{k} \leq \frac{1}{k!} \leq \frac{1}{2^{k-1}}$$ (c)Show that for $n \in \mathbb{N}:$ $$2 \leq (1+ \frac{1}{n})^n &lt; 3$$ Use (a) and the binomial theorem for (b). Use the following for (c): $$\sum_{j=0}^{n}q^j=\frac{1-q^{n+1}}{1-q}$$  I've already proven (a) thanks to @trancelocation. Now I got stuck on (b) and (c). (b) I don't know how to use (a) and the binomial theorem here. $\frac{1}{n^k}  \binom{n}{k} \leq \frac{1}{k!} \leq \frac{1}{2^{k-1}} \Leftrightarrow \frac{n!}{n^k*(n-k)!* k!}\leq \frac{1}{k!} \leq \frac{1}{2^{k-1}} \Leftrightarrow \frac{n!}{n^k*(n-k)!}\leq 1 \leq \frac{k!}{2^{k-1}}$ Now I have to show: $\frac{n!}{n^k*(n-k)!}\leq 1$ and $1 \leq \frac{k!}{2^{k-1}}$ $\frac{n!}{n^k*(n-k)!} = \frac{n*(n-1)*...*(n-k+1)*[(n-k)*...*1]}{n^k*[(n-k)*...*1]}=\frac{n*...*(n-k+1)}{n^k}=\frac{n}{n}*\frac{n-1}{n}*...*\frac{n-k+1}{n} \leq 1$. The first factor is $1$. Starting from the second factor their value gets smaller, so the entire term gets smaller then $1$. I tried to do it like this: for $i=0, ..., n$:  $\frac{n-i}{n} \leq 1 \Leftrightarrow n-i \leq n \Leftrightarrow 0 \leq i$ For the second part: $\frac{k!}{2^{k-1}} \geq 1 \Leftrightarrow \frac{k!}{2^k*2} \geq 1 \Leftrightarrow \frac{k!}{2^k} \geq 2\Leftrightarrow \frac{k}{2}*\frac{k-1}{2}*\frac{k-2}{2}*...*\frac{1}{2} \geq 2$ So for $\frac{k-j}{2} \geq 2$, with $j=0, ..., k-1$ $\frac{k-j}{2} \geq 2 \Leftrightarrow k-j \geq 4$ This doesn't seem to help at all. (c)$2 \leq (1+ \frac{1}{n})^n &lt; 3$ I didn't know how to start at c) at all. The provided formula doesn't help me.
67555
I am given that $${n \choose a}={n \choose b}$$ I need to show that $n=a+b$ or $a=b$.  Now I am left with $a!(n-a)!=b!(n-b)!$.  I don't know what to do next. I checked there on google to solve the above, it says:- "On comparing, we get the required result." But it makes no sense to me.
9612
We have to rotate a rectangle along the circle at an angle $\theta$ and then move the rectangle along the tangent for a distance 'lateral offset' as given in the following image  I calculated the lateral offset factors to be applied on the rectangle (after rotating it) coordinates as given below. Reference -          Slope = tan(θ);         SlopeLateralOffset = -1.0 / Slope;         LateralOffsetTranslationFactor = LateralOffset / sqrt(1.0 + SlopeLateralOffset  * SlopeLateralOffset);          LateralOffsetTranslationX = LateralOffsetTranslationFactor ;         LateralOffsetTranslationY = SlopeLateralOffset  * LateralOffsetTranslationFactor ;  Is the logic that I applied correct? I think that the formula SlopeLateralOffset = -1.0 / Slope, will not work always as I cannot form a right triangle with radius r and lateral offset What changes should I make it to work? Can you please help me?
73673
You are given four points (on a Euclidian plane) that make up the corners of a square. You may change the positions of the points by a sequence of moves. Each move changes the position of one point, say p, to a new location, say p', by "skipping over" one of the other 3 points. More precisely, p skips over a point q if it is moved to the diametrically opposed side of q. In other words, a move from p to p' is allowed if there exists a point q such that $q = (p + p') / 2$.  Find a sequence of moves that results in a larger square. Or, if no such sequence is possible, give a proof of why it isn't possible. (The new square need not be oriented the same way as the original square. For example, the larger square may be turned 45 degrees from the original, and the larger square may have the points in a different order from in the original square.)  I've tried many different combinations and still haven't gotten anything more than 3 points in the right position and one point completely off. Is this possible?
98478
I have a programming problem and I've forgotten some of my math rectangle formulas. I have to make a program that creates a rectangle that isn't parallel to the X and Y axis, I assume this means a diagonal rectangle in the first quadrant of a Cartesian graph.  My inputs are the following:    coordinates of the vertex closest to the X axis and the origin angle the long side makes with relation to the X axis. area of the rectangle length of the long side   The outputs expected are the following:   length of the short side,  coordinates of all the rest of the rectangle's vertices.   Given these input how do I calculate the output mentioned?
108970
I am revising for an exam and have just realised that the euclidean algorithm questions in past exams are much harder than in homeworks! So i need some help please.  I have a question here, i already have the solution to it but i dont understand it that well, so i would like someone to please explain what is going on ect, and dont worry about avoiding giving me the full solution as i already have it so a full solution would be good!  Here is the question:  Use euclids Algorithm to find $gcd(172,20)$.  Hence solve $172a + 20b = 1000$ giving all solutions in terms of parameter t. List any solutions for which $a,b&gt;0$.  So it wants the answers in terms of parameter t, which is something i am not familiar with and cannot find in my books.   I have the solution in the solutions paper as   $$172= 8\times20+12 $$ $$20=1\times 12 + 8$$ $$12=1 \times 8 +4$$ $$8=2\times 4+0$$  thus $gcd(172,20)=4$  then   $$4= 12-8=12-(20-12)=2\times 12 - 20$$  $$=2\times(172-8\times20)-20$$  $$=2 \times 172 -17 \times20$$   multiplying by 250 we get   $$1000=500 \times 172 -4250 \times 20$$  so one solution is $a=500, b=-4250$  So, I understand everything up until here!!  Now it says,  Generally, $$a=500-t \frac{20}{4}=500-t$$            $$b=-4250+t \frac{172}{4}=-4250+43t$$  To have $a,b&gt;0$, we need $$500-5t&gt;0, (100&gt;t)$$ to have $a&gt;b$ we need $$-4250+43t&gt;0,(t&gt;98.8)$$$  Need $t=99$, $a=5$, $b=7$  Could someone please help explain what is going on in this last bit of working out? I am pretty good at euclidean algorithm usually but havent seen this type before an am panicking  bit as so close to exam! Any help appreciated how to find t and the other positive solutions would be great!  Please could you show me exactly how it is done here as my lecturer is quite strict with us using his exact methods. Many thanks
96224
I tried to prove that the area of a rectangle is $ab$ given side lengths $a$ and $b$.  The best I can do is the assume the area of a $1\times1$ square is $1$. Then not the number of $1\times1$ squares that fit in an $a\times b$ rectangle is $ab$. Therefore area is $a\cdot b$. This does not seem rigorous however.
3540
I don't know what to do with this. Nothing works. I hope somebody can help me to find a decision $$n!\leq {\left(\frac{n+1}{2}\right)}^{n}$$
26108
If I was at point $A$ and I wanted to walk directly to point $B$, I would have to walk half way to point $B$, but before that I would have to walk half way to halfway to halfway to point $B$ and half of that again and so on and so fourth. if I halved this distance an infinite amount of times then there would be an infinite amount of actions I would need to perform in order to cross from $A$ to $B$. Therefore, if each action took any quantity of time at all, then it would take me an infinite amount of time to cross from $A$ to $B$, even if $A$ and $B$ were only a few centimetres apart!   Because I am not infinitely old and I can move, there must be a flaw in this logic. Where is it?
122842
I perfectly understand the permutations formula i.e. if you have $n$ things how many ways can you rearrange it if taken $k$ at a time (or if you have $k$ slots)? So you draw the following tree. And the formula comes out naturally. $$^nP_k = \frac{n!}{(n-k)!}$$  I get that it is including all of the duplicates, too. And to get rid of them we use the combinations formula: $$^nC_k = \frac{n!}{k!(n-k)!}$$ Can somebody explain why we have to divide by the &quot;number of ways to re-arrange the slots&quot; to get rid of the duplicates? I understand that it works and the division is used to scale a number down but I am not getting that &quot;aha moment&quot; of why we do it. Can somebody explain it?
21165
How to simplify this best  $(a_1 + a_2 +a_3+... +a_n)^m$ for  $m=n, m&lt;n, m&gt;n$  I could only get  $\sum_{i=0}^{m}\binom{m}{i}a_i^i\sum_{j=0}^{m-i}\binom{m-i}{j}a_j ... $
92530
Consider the red path from A that zigzags to B, which takes $n$ even steps of length $w$. The path length of the route $P_n$ will be equal to:  $ P_n = P_x + P_y = \frac{n}{2}\times w + \frac{n}{2}\times w = n \times w $  But $\frac{n}{2}\times w = 1$ beacuse it is the length of one of the sides of the triangle so:  $P_n = 2$  Which will be true no matter how many steps you take. However in the limit $n \to \infty, w \to 0$ the parth length $P_\infty$ suddenly becomes:  $P_\infty = \sqrt{1^2 + 1^2} =  \sqrt{2}$  Due to Pythagoras. Why is this the case? It seems the path length suddenly decreases by 0.59!
57035
If I want to find how many possible ways there are to choose k out of n elements I know you can use the simple formula below:  $$ \binom{n}{k} = \frac{n! }{ k!(n-k)! } .$$  What if I want to go the other way around though?  That is, I know I want to have $X$ possible combinations, and I want to find all the various pairs of $n$ and $k$ that will give me that number of combinations.  For example, if the number of combinations I want is $3$, I want a formula/method to find that all the pairs that will result in that number of combinations are $(3,1)$ and $(3,2)$  I know I could test all the possible pairs, but this would be impractical for large numbers.  But perhaps there's no easier way of doing this then the brute force approach?
33362
I have four cities in $A=(0,0),B=(1,0),C=(1,1),D=(0,1)$. I am asked to build the shortest motorway to connect the cities. How can I do that?  I was thinking that first I need some compactness argument to prove that there really is the shortest way to connect cities. Or if one can prove that one can travel via finitely many line segments, the existence of minimal way migth be possible to prove via some olympiad-level inequalities.  Then I think I need to find points $E,F$ such that $\angle AEB=\angle CFD=120^\circ$. But I think it is difficult to prove that such a configuration is really a minimal.  Could anyone give hints/solution how to fill the details?  Bonus: Has anyone generalized this problem to $n$ cities in a regular $n$-gon?
47906
Suppose we have two numbers A and B. We need to find a multiple of A such the A % B is maximized One such solution can be to traverse B multiples of A? is there any other soltuion?(In case B is huge)  Suppose A=3 and B=10. then 10 multiples of 3 are 3,6,9.......30.   Max remainder will be 9%10=9.  Suppose A=14 and B=11 then 11 multiples of 14 are 14,28..........140,154.  Then max remainder 98%11=10 .  Is there any more effective way to find this?
18569
I have come across numerous questions where I am given the picture such as the above one been asked "how many rectangles are there?". I have even come across some slightly different images that instead of rectangles you are supposed to find the number of triangles. Well, I was thinking whether there is any formula or strategy that is used to solve these problems without having to manually count every shape.  Help would be appreciated.  Thank you :)