In 20 over game.... first team scored 152 all out in 16.2 Over and second team scored 153/6 in 17 over.what is the run rate?
I have some doubts regarding net run rate in the ongoing ICC world twenty 20 2014 event.It takes a while to read so please don't hesitate and give me a solution regarding this.    My doubt is that in match 12 of world twenty 20 2014(In Group Stage Last Match on march 21 2014) () Wesley Barresi of Netherlands hit a six when the team score was 187 while chasing Ireland score of 189, That take Netherlands score to 193.  If that is the case in match 17 between India and West Indies in match of super10 group (in ICC world twenty 20 2014 on march 23 2014) () while Suresh Raina (India) hit a four when the team score was 129 while chasing West Indies score of 129,That takes India Only To 130. why Only One run is added to the score in this match(match 17 on march 23 2014) while 6 runs are added to the score in match 12 on march 21 2014.  I don't understand the strategy behind this.Because if India's Score is taken as 133 then India's net run rate as of now(24 march 2014) will be 0.442(Runs Scored For 264 in 38.1 overs ) But Now it is only 0.383(Runs Scored For 261 in 38.1 overs ).As the net run rate plays an important role in the tournaments like this i am not very happy with this.  As being an Indian Fan now a days I look a lot in to net run rate as it has eliminated India in many tournaments including last ICC world twenty 20 2012.  Is There Any One who can clarify my doubts regarding the net run rate as i want to be sure of what was happening?
I'd like to download automatically tons of papers listed on .bib file so I'm trying to find a way to facilitate the process by, e.g. I import or input to a tool a certain .bib file which contains a list of literature, the output is the literature's .pdf file. Any idea to do so? Proof of the duality of the dominance order on partitions Could anyone provide me with a nice proof that the  $\leq$ on partitions of an integer $n$ satisfies the following: if $\lambda, \tau$ are 2 partitions of $n$, then $\lambda \leq \tau \Longleftrightarrow \tau &#39; \leq \lambda &#39;$, where $\lambda&#39;$ is the conjugate partition of $\lambda$ (i.e. the transpose of the set of 'dots' which $\lambda$ represents).   I feel like there must be a nice clever and concise/intuitive proof of this, but all I can come up with is an ugly brute force approach based on the definition of sums of the components $\lambda_i$. Could anyone suggest a particularly nice way to obtain this result? Many thanks. LCM is associative: $\ \text{lcm}(\text{lcm}(a,b),c)=\text{lcm}(a,\text{lcm}(b,c))$ Definition: The least common multiple of $a.b\in\Bbb{Z}$ is the smallest  $k\in\Bbb{N}$ that is divisible by both a and b.         Proposition: $\text{lcm}(\text{lcm}(a,b),c)=\text{lcm}(a,\text{lcm}(b,c))\tag{2}$    Proof: We want to show $\text{lcm}(\text{lcm}(a,b),c)\mid\text{lcm}(a,\text{lcm}(b,c))$ and $\text{lcm}(a,\text{lcm}(b,c))\mid \text{lcm}(\text{lcm}(a,b),c)$ such that $\text{lcm}(\text{lcm}(a,b),c)=\text{lcm}(a,\text{lcm}(b,c))$.  Let $k=\text{lcm}(\text{lcm}(a,b),c)$, then $c\mid k$ and $\text{lcm}(a,b)\mid k$, but if we let $l=\text{lcm}(a,b)$, then $a\mid l\,\,\text{and}\,\, b\mid l$. Therefore $c\mid k\,\,,a\mid k\,\,\text{and}\,\, b\mid k$, but then $\text{lcm}(b,c)\mid k$. Thus $\text{lcm}(a,\text{lcm}(b,c))\mid k$      Question: Does this makes sense, ifso can I do the same for $\text{lcm}(\text{lcm}(a,b),c)\mid \text{lcm}(a,\text{lcm}(b,c))$? 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