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

$\begin{array}{l}Totalnumberofwordsin\text{'}TRIANGLE\text{'}=8\\ Outof5areconsonantsand3arevowels\\ Ifvowelsarenottogether,takenwehavethefollowingarrangement\\ VUnsupported\; ?CUnsupported\; ?VUnsupported\; ?CUnsupported\; ?VUnsupported\; ?CUnsupported\; ?VUnsupported\; ?CUnsupported\; ?VUnsupported\; ?CUnsupported\; ?V\\ consonantcanbearrangedin5!=120ways\\ Voweloccupy6places\\ \therefore 3vowelscanbearrangedin6places={}_6{P}_3\\ =\frac{6!}{\left(6-3\right)!}=\frac{6!}{3!}=120ways\\ So,thetotalarrangement=120\times 120=14400ways\\ Here,therequiredarrangement=14400ways.\end{array}$

Adani Skill Development application forms are out for the academic year 2025-26 for all certificate courses. To get the application form candidates are required to first fill out the Request Call Back form before the last date. Candidates can apply online through the official website of the college. The admission to the course is direct. Additionally, candidates must have to fill out the admission forms to secure their seat.

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

$\begin{array}{l}Totalnumberofthings=n\\ 3thingsmustbetogether\\ \therefore thenumberofremainingthings=n-3\\ Numberofthingstobeselected=r\\ Outofr,3arealwaystogether\\ \therefore Numberofwaysofselection={}_{n-3}{C}_{r-2}\\ Nowpermutationof3thingswhicharealwaystogether=3!\\ Numberofpermutationof\left(r-2\right)things=\left(r-2\right)!\\ \therefore Totalnumberofarrangements={}_{n-3}{C}_{r-2}\times \left(r-2\right)!\times 3!\\ Hence,therequiredarrangements={}_{n-3}{C}_{r-2}\times \left(r-2\right)!\times 3!.\end{array}$

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

$\begin{array}{l}GeneralTerm{T}_{r+1}={}_n{C}_r{x}^{n-r}{y}^r\\ ={}_{15}{C}_r{\left(\frac{3x^2}{2}\right)}^{15-r}{\left(-\frac{1}{3x}\right)}^r={}_{15}{C}_r{\left(\frac{3}{2}\right)}^{15-r}{\left(x\right)}^{30-2r}{\left(-\frac{1}{3}\right)}^r\frac{1}{x^r}\\ ={}_{15}{C}_r{\left(\frac{3}{2}\right)}^{15-r}{\left(x\right)}^{30-2r-r}{\left(-1\right)}^r\frac{1}{{\left(3\right)}^r}\\ ={}_{15}{C}_r{\left(\frac{3}{2}\right)}^{15-r}{\left(x\right)}^{30-3r}{\left(-1\right)}^r\frac{1}{{\left(3\right)}^r}\\ forgettingthetermindependentofx,\\ 30-3r=0\Rightarrow r=10\\ Onputtingthevalueofrintheaboveexpression,weget\\ ={}_{15}{C}_{10}{\left(\frac{3}{2}\right)}^{15-10}{\left(-1\right)}^{10}\frac{1}{{\left(3\right)}^{10}}={}_{15}{C}_{10}\frac{{\left(3\right)}^5}{{\left(2\right)}^5}.\frac{1}{{\left(3\right)}^{10}}\\ ={}_{15}{C}_{10}.\frac{1}{{\left(2\right)}^5.{\left(3\right)}^5}={}_{15}{C}_{10}{\left(\frac{1}{6}\right)}^5\\ Hence,therequiredterm={}_{15}{C}_{10}{\left(\frac{1}{6}\right)}^5\end{array}$

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

$\begin{array}{l}Giventhatbagcontains5blackand6redballs.\\ Numberofwaysofselecting2blackballsoutof5blackballs={}_5{C}_2\\ andnumberofwaysofselecting3redballsoutof6redballs={}_6{C}_3\\ \therefore Totalnumberofwaysofselecting2blackand3redballs={}_5{C}_2\times {}_6{C}_3\\ =\frac{5.4}{2.1}\times \frac{6.5.4}{3.2.1}=10\times 20=200ways\\ Hence,therequiredwaysofselectingtheballs=200.\end{array}$

During the IIT Bhilai placements 2024, the overall average package offered stood at INR 13.02 LPA. The institute has released the average packages for BTech, M.Tech and MSc programs as well. Check out the table presented below for the average packages offered during placements held between 2022-2024:

The courses available at Adani Skill Development are as follows:

Field Technician – Networking & Storage

Field Technician – Computing Peripherals

Mobile Phone Hardware Repair Technician

Agarbatti Making
Mining Mechanic – Fitter

Candidates can choose from any of the courses available based on their interests and career goals. 

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

$\begin{array}{l}Totalnumberofpersons=12\\ Numberofpersonstobeselected=5\\ Wehave26Englishalphabetand10digits\left(0to9\right)\\ Since,itisgiventhateachplatecontains2differentlettersfollowedby3differentdigits.\\ \therefore Numberofarrangementof26letterstaken2atatime={}_{26}{P}_2=\frac{26!}{\left(26-2\right)!}=\frac{26!}{24!}\\ =\frac{26.25.24!}{24!}=650\\ Threedigitnumbercanbeformedoutof10digit={}_{10}{P}_3=\frac{10!}{7!}\\ =\frac{10.9.8.7!}{7!}=720\\ \therefore Totaslnumberoflicenseplates=650\times 720=468000\\ Hence,therequirednumberofplates=468000.\end{array}$

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

$\begin{array}{l}Totalnumberofpersons=12\\ Numberofpersonstobeselected=5\\ Outof5,thereisachairperson\\ \therefore Numberofwaysofselectingachairperson={}_{12}{C}_1=12\\ Numberofwaysofselectingother4numbersoutofremaining11persons={}_{11}{C}_4\\ \therefore Totalnumberofways={}_{12}{C}_1\times {}_{11}{C}_4\\ =12\times \frac{11.10.9.8}{4.3.2.1}=12\times 330=3960\\ Hence,therequirednumberofways=3960.\end{array}$

Yes, a candidate can apply for admission in Indian Institute for Aeronautical Engineering and Information Technology (IIAEIT) even if their class 12 results are pending. The institute may provide them provisional admission, and they can appear for the AERO-CET exam. However, for admission confirmation marksheets are required.

SBI Clerk prelims exam analysis 2026 will be available here after the conduction of an exam. The exam analysis is made available after the completion of each exam shift. SBI Clerk paper analysis highlight the section-wise and overall difficulty level of the exam. Going through the exam analysis candidates can know the topics from where questions are asked in the exam and their difficulty levels. 

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

$\begin{array}{l}Totalnumberofpersons=8\\ Numberofpersonstobeselected=6\\ ConditionisthatifAischoosen,Bmustbechoosen\\ CaseI:WhenAischoosen,Bmustbechoosen\\ Numberofways={}_6{C}_4\left[?AandBaresettobechoosen\right]\\ CaseII:WhenAisnotchoosen,thenBmaybechoosen\\ Numberofways={}_7{C}_6\\ So,thetotalnumberofways={}_6{C}_4+{}_7{C}_6\left[?Therearetwocases\right]\\ ={}_6{C}_2+{}_7{C}_1\left[?{}_n{C}_r={}_n{C}_{n-r}\right]\\ =\frac{6.5}{2.1}+7=15+7=22ways\\ Hence,therequirednumberofways=22.\end{array}$