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$Finding largest prime number which divide $\displaystyle \binom{2000}{1000}$ $We can write $\displaystyle \binom{2000}{1000} = \frac{2000!}{1000!\cdot 1000!} = \frac{1001 \cdot 1002 \cdot 1003\cdots \cdots 2000}{1000!}$ i did not understand how to solve further, please explain me

$There are four machines and it is known that exactly two of them are $ $faulty. They are tested, one by one, in a random order till both the faulty $ $machines are identified. Then the probability that only two tests are$ $need is$ $What i have Try:$ $Let $A$ be the event in...

$Total number of positive integer ordered pair of $\binom{a}{b} = 120$ $Using $\binom{a}{b} = 120 = \binom{120}{1} = \binom{120}{119}$. So $(a,b) = (120,1)\;,(120,119)$ $And $\binom{a}{b}$ is $\max$, when $b=\frac{a}{2}$ or $b=\frac{a+1}{2}$ $So must have $b\leq \frac{a}{2}$ or $b \leq...

\displaystyle \bigg\lfloor\frac{2017!}{1!+2!+3!+\cdots \cdots +2016!}\bigg\rfloor $where $\lfloor x \rfloor $ is a floor of $x$ $For upper bound $1!+2!+3!+\cdots \cdots +2016!=2016(1!+2!+3!+\cdots +2016!)$ 2016(1!+2!+3!+\cdots \cdots +2016!)>2016(2016!+2015!)=2017! \displaystyle...

my bad omegadot you are Right., I have got it, Thanks

Sorry friends actually original question as $If $I = \int^{1}_{0}x^{\frac{5}{2}}(1-x)^{\frac{7}{2}}dx$ and $J = \int^{1}_{0}\frac{x^{\frac{5}{2}}(1-x)^{\frac{7}{2}}}{(3+x)^8}dx$ . Then $\displaystyle \frac{I}{J}$ is

$Evaluation of sum of series $\lim_{n\rightarrow \infty}\sum^{n}_{k=1}\frac{1}{(k-1)\sqrt{k}+k\sqrt{k+1}}$

$Using $\sin^2 x+\cos^2 x = 1\Rightarrow \sin^2 x+\cos^2 x+(-1) = 0$ $Now if $a+b+c=0\;,$ Then $a^3+b^3+c^3=3abc$ $So $(\sin^2 x)^3+(\cos^2 x)^3+(-1)^3=-3\sin^2 x\cdot \cos^2 x$ $So $\sin^6 x+\cos^6 x= \frac{1}{4}\left[4-3(\sin 2x)^2\right] = \frac{1}{4}\left[1+3\cos^2 (2x)\right].$

$If $I = \int^{1}_{0}x^{\frac{5}{2}}(1-x)^{\frac{7}{2}}dx$ and $J = \int^{1}_{0}\frac{x^{\frac{3}{2}}(1-x)^{\frac{7}{2}}}{(3+x)^8}dx$ . Then $\displaystyle \frac{I}{J}$ is

Sorry pikachu97 , i have edited my post.

Re: HSC 4U Integration Marathon 2017 $Evaluation of $\int \frac{x\cos x+1}{\sqrt{2x^3 \cdot e^{\sin x}+x^2}}dx$ $Evaluation of $\int^{\infty}_{-\infty}\frac{\ln(1+16x^2)}{1+25x^2}dx$

$Let $z_{1},z_{2}$ be two roots of $az^2+bz+c=0.$ So $z_{1}+z_{2} = -\frac{b}{a}\;,z_{1}\cdot z_{2} = \frac{c}{a}$ $Given $|z_{1}| = |z_{2}| = 1$ and $|a| = |b| = |c| = 1$ $So $|z_{1}+z_{2}|^2 = 1\Rightarrow (z_{1}+z_{2})\cdot (\bar{z_{1}}+\bar{z_{2}}) = 1\Rightarrow...

$If $a,b,c$ are three distinct non zero complex numbers such that $ |a| = |b| = |c|$ and the equation $az^2+bz+c=0$ has a roots whose$ $ modulus is $1\;,$ Then relation between $a,b,c$ is$

Re: HSC 4U Integration Marathon 2017 $Evaluation of $\int\frac{3+4\cos x}{(4+3\cos x)^2}dx$ $Evaluation of $\int^{\frac{\pi}{4}}_{0}(\cos 2x)^{\frac{3}{2}}\cdot \cos xdx$

$Find all parameter $a$ for which the equation $(a-1)4^x-4\cdot 2^x+a+2=0$ $has at least one real solution$

Re: MX2 2016 Integration Marathon $Evaluation of $\int^{\infty}_{0}\lfloor ne^{-x} \rfloor dx,$ where $n\in \mathbb{N},n>1$ $and $\lfloor x \rfloor$ represent floor of $x$

Re: MX2 2016 Integration Marathon $(1) Evaluation of $\int\frac{1}{\sec x+\csc x}dx$ $(2) Evaluation of $\int^{\frac{\pi}{2}}_{0}\frac{\cos^6 x}{\sin x+\cos x}dx$ $(3) Evaluation of $\int^{\frac{\pi}{4}}_{0}\frac{\sin^2 x\cos^2 x}{\sin^3 x+\cos^3 x}dx$

$Evaluatiuon of $\sum_{0\leq i <j \leq n}j\binom{n}{i}$ and $\sum_{0\leq i \leq j \leq n}\binom{n}{i}\binom{n}{j}$

$If the coefficients of bi-quadratic equation are all distinct and belong $ $to the set $\{-9,-5,3,4,7\}.$ Then prove that the bi-quadratic equation has$ $at least two real roots$

Thanks KingOfActing.