Class 11th

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l}Totalnumberofstudentsineachclass=20\\ Wehavetoselectatleast5studentsfromeachclass.\\ Wehavethefollowingcases.\\ \left(i\right)5studentsfromXIclassand6studentsfromXIIclass\\ \left(ii\right)6studentsfromXIclassand5studentsfromXIIclass\\ So,numberofwaysofselectionofateamof11players\\ ={}_{20}{C}_5*{}_{20}{C}_6+{}_{20}{C}_6*{}_{20}{C}_5=2\left[{}_{20}{C}_5*{}_{20}{C}_6\right]\\ Hence,therequiredwaysofselection=2\left[{}_{20}{C}_5*{}_{20}{C}_6\right]\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Asperthegiveninformationwehavethefollowingdiagram\\ Startingpoint=S\\ Distancetravelledtobringthefirstpotato=24+24=48m\\ Distancetravelledtobringthesecondpotato=2\left(24+4\right)=56m\\ Distancetravelledtobringthethirdpotato=2\left(24+4+4\right)=64m\\ Therfore,theserieswillbe=48,56,64,\dots \\ whichan\text{?}\text{?}A.Pinwhicha=48,d=56-48=8\\ Wehavetofindthetotaldistancetobringallthepotatoesback,so,n=20\\ \therefore S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\\ \Rightarrow S_{20}=\frac{20}{2}\left[2*48+\left(20-1\right)8\right]=10\left[96+152\right]\\ =10*248=2480m\\ Hence,therequireddistance=2480m\end{array}$

This is a Long Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l}Thegivenexpressionis{\left(x^2+\frac{1}{x}\right)}^{2n}\\ GeneralTerm T_{r+1}={}_n{C}_r{x}^{n-r}{y}^r\\ ={}_{2n}{C}_r{\left(x^2\right)}^{2n-r}{\left(\frac{1}{x}\right)}^r={}_{2n}{C}_r{\left(x\right)}^{4n-2r}.\frac{1}{x^r}\\ ={}_{2n}{C}_r{\left(x\right)}^{4n-2r-r}={}_{2n}{C}_r{\left(x\right)}^{4n-3r}\\ If{x}^{p}occursin{\left(x^2+\frac{1}{x}\right)}^{2n}\\ then4n-3r=p\Rightarrow 3r=4n-p\\ \Rightarrow r=\frac{4n-p}{3}\\ \therefore Coefficientof{x}^p={}_{2n}{C}_r={}_{2n}{C}_{\frac{4n-p}{3}}\\ =\frac{\left(2n\right)!}{\left(\frac{4n-p}{3}\right)!\left(2n-\frac{4n-p}{3}\right)!}=\frac{\left(2n\right)!}{\left(\frac{4n-p}{3}\right)!\left(\frac{6n-4n+p}{3}\right)!}\\ =\frac{\left(2n\right)!}{\left(\frac{4n-p}{3}\right)!\left(\frac{2n+p}{3}\right)!}\\ Hence,coefficientof{x}^p=\frac{\left(2n\right)!}{\left(\frac{4n-p}{3}\right)!\left(\frac{2n+p}{3}\right)!}\end{array}$

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l}Giventhatthetotalnumberofplayers=16\\ Wehavetoselect11playersoutof16players.\\ \left(i\right)If2playersareincluded,thennumberofwaysofselection={}_{16-2}{C}_{11-2}={}_{14}{C}_9\\ \left(ii\right)If2playersareexcluded,thennumberofwaysofselection={}_{16-2}{C}_{11}={}_{14}{C}_{11}\\ Hence,therequirednumberofwaysofselectionare\\ \left(i\right){}_{14}{C}_9\left(ii\right){}_{14}{C}_{11}\end{array}$

This is a Long Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l}\left(i\right)Thegivenexpressionis{\left(x+a\right)}^n\\ {\left(x+a\right)}^n={}_n{C}_0{x}^n{a}^0+{}_n{C}_1{x}^{n-1}a+{}_n{C}_2{x}^{n-2}{a}^2+{}_n{C}_3{x}^{n-3}{a}^3+\dots +{}_n{C}_n{a}^n\\ Sumofoddterms,\\ O={}_n{C}_0{x}^n+{}_n{C}_2{x}^{n-2}{a}^2+{}_n{C}_4{x}^{n-4}{a}^4+\dots \\ andthesumofeventerms,\\ E={}_n{C}_1{x}^{n-1}a+{}_n{C}_3{x}^{n-3}{a}^3+{}_n{C}_5{x}^{n-5}{a}^5+\dots \\ Now{\left(x+a\right)}^n=O+E\dots \left(i\right)\\ Similarly,{\left(x-a\right)}^n=O-E\dots \left(ii\right)\\ Multiplyingeqn.\left(i\right)andeqn.\left(ii\right),weget\\ {\left(x+a\right)}^n{\left(x-a\right)}^n=\left(O+E\right)\left(O-E\right)\\ {\left(x^2-a^2\right)}^n=O^2-E^2\\ Hence,O^2-E^2={\left(x^2-a^2\right)}^n\\ \left(ii\right)4OE={\left(O+E\right)}^2-{\left(O-E\right)}^2\\ ={\left[{\left(x+a\right)}^n\right]}^2-{\left[{\left(x-a\right)}^n\right]}^2\\ ={\left[x+a\right]}^{2n}-{\left[x-a\right]}^{2n}\\ Hence,4OE={\left(x+a\right)}^{2n}-{\left(x-a\right)}^{2n}\end{array}$

This is a Long Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l}Thegivenexpressionis{\left(+\frac{1}{}\right)}^n\\ ={\left(2^{1/3}+\frac{1}{3^{1/3}}\right)}^n\\ GeneralTerm T_{r+1}={}_n{C}_r{x}^{n-r}{y}^r\\ T_7=T_{6+1}={}_n{C}_6{\left(2^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}^{n-6}{\left(\frac{1}{3^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\right)}^6\\ ={}_n{C}_6{\left(2\right)}^{\raisebox{1ex}{$n-6$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}.\left(\frac{1}{3^2}\right)={}_n{C}_6{\left(2\right)}^{\raisebox{1ex}{$n-6$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}.{\left(3\right)}^{-2}\\ 7thtermfromtheend={\left(n-7+2\right)}^{th}termfromthebeginning\\ ={\left(n-5\right)}^{th}termfromthebeginning\\ So,T_{n-6+1}={}_n{C}_{n-6}{\left(2^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}^{n-n+6}{\left(\frac{1}{3^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\right)}^{n-6}\\ ={}_n{C}_{n-6}{\left(2\right)}^2.\left(\frac{1}{3^{\raisebox{1ex}{$n-6$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\right)={}_n{C}_{n-6}{\left(2\right)}^2.{\left(3\right)}^{\raisebox{1ex}{$6-n$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\\ Accordingtothequestion,weget\\ \frac{{}_n{C}_6{\left(2\right)}^{\raisebox{1ex}{$n-6$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}.{\left(3\right)}^{-2}}{{}_n{C}_{n-6}{\left(2\right)}^2.{\left(3\right)}^{\raisebox{1ex}{$6-n$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}=\frac{1}{6}\\ \Rightarrow \frac{{}_n{C}_{n-6}{\left(2\right)}^{\raisebox{1ex}{$n-6$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}.{\left(3\right)}^{-2}}{{}_n{C}_{n-6}{\left(2\right)}^2.{\left(3\right)}^{\raisebox{1ex}{$6-n$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}=\frac{1}{6}\Rightarrow {\left(2\right)}^{\frac{n-6}{3}-2}.{\left(3\right)}^{-2-\frac{6-n}{3}}=\frac{1}{6}\\ \Rightarrow {\left(2\right)}^{\frac{n-6-6}{3}}.{\left(3\right)}^{\frac{-6-6+n}{3}}=\frac{1}{6}\Rightarrow {\left(2\right)}^{\frac{n-12}{3}}.{\left(3\right)}^{\frac{n-12}{3}}={\left(6\right)}^{-1}\\ \Rightarrow {\left(6\right)}^{\frac{n-12}{3}}={\left(6\right)}^{-1}\Rightarrow \frac{n-12}{3}=-1\\ \Rightarrow n-12=-3\Rightarrow n=12-3=9\\ Hence,therequiredvalueofnis9.\end{array}$