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Author: Tinku Tara

64-points-are-in-a-plane-x-y-x-0-1-2-7-y-0-1-2-7-4-points-are-chosen-at-random-What-is-the-proabability-the-lines-connecting-them-do-not-form-a-square-or-rectangle-

Question Number 2771 by prakash jain last updated on 26/Nov/15 $$\mathrm{64}\:\mathrm{points}\:\mathrm{are}\:\mathrm{in}\:\mathrm{a}\:\mathrm{plane}: \\ $$$$\left({x},{y}\right),\:{x}\in\left\{\mathrm{0},\mathrm{1},\mathrm{2},…,\mathrm{7}\right\},\:{y}\in\left\{\mathrm{0},\mathrm{1},\mathrm{2},…,\mathrm{7}\right\} \\ $$$$\mathrm{4}\:\mathrm{points}\:\mathrm{are}\:\mathrm{chosen}\:\mathrm{at}\:\mathrm{random}. \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{proabability}\:\mathrm{the}\:\mathrm{lines}\:\mathrm{connecting} \\ $$$$\mathrm{them}\:\mathrm{do}\:\mathrm{not}\:\mathrm{form}\:\mathrm{a}\:\mathrm{square}\:\mathrm{or}\:\mathrm{rectangle}? \\ $$ Answered by Rasheed Soomro…

Question-133840

Question Number 133840 by Algoritm last updated on 24/Feb/21 Answered by Dwaipayan Shikari last updated on 24/Feb/21 $$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{{x}^{{x}} \Gamma\left({x}+\mathrm{1}\right)−\mathrm{4}\Gamma\left(\mathrm{3}{x}−\mathrm{3}\right)}{\mathrm{3}\Gamma\left({x}+\mathrm{1}\right)−\mathrm{6}}=\frac{{x}^{{x}} \left({logx}+\mathrm{1}\right)\Gamma\left({x}+\mathrm{1}\right)+\Gamma'\left({x}+\mathrm{1}\right){x}^{{x}} −\mathrm{12}\Gamma'\left(\mathrm{3}{x}−\mathrm{3}\right)}{\mathrm{3}\Gamma'\left({x}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{8}\left({log}\left(\mathrm{2}\right)+\mathrm{1}\right)+\mathrm{4}\Gamma'\left(\mathrm{3}\right)−\mathrm{12}\Gamma'\left(\mathrm{3}\right)}{\mathrm{3}\Gamma'\left(\mathrm{3}\right)}=\frac{\mathrm{8}\left({log}\left(\mathrm{2}\right)+\mathrm{1}\right)}{\mathrm{6}\left(\frac{\mathrm{3}}{\mathrm{2}}−\gamma\right)}−\frac{\mathrm{8}}{\mathrm{3}} \\…

Without-using-induction-or-arithmatic-series-concept-prove-the-following-1-2-3-n-n-n-1-2-

Question Number 2762 by Rasheed Soomro last updated on 26/Nov/15 $${Without}\:{using}\:\underset{−} {{induction}}\:{o}\underset{−} {{r}\:\:{arithmatic}\:{series}−{concept}\:\:\:} \\ $$$$\:{prove}\:{the}\:{following}: \\ $$$$\mathrm{1}+\mathrm{2}+\mathrm{3}+…+{n}=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$ Answered by prakash jain last updated…