It's a Question of your Career!

Hi

This is very difficult to predict because cut off keeps changing every year depending upon a number of factors like

--------) number of students that appeared in the WBJEE exam

--------)difficulty level of the WBJEE examination

--------)Kind of marks obtained by the students in the WBJEE exam

--------) number

Hello Aspirant,

See even if you are from Tamil Nadu state board then also you have to fill only the class 12th roll number in the 12th std. roll no column. You will find you class 1th roll number mentioned on top of your Admit card or if not then

Dear student,

With your TS EAMCET Rank it is difficult to get free seat in G. Narayanmma Institute of Technology and Science, Rayadurg that too Bachelor of Technology (B.Tech) Computer Science and Engineering. Previous year's closing rank for B.Tech Computer Science and Engineering in this college for BC- E category

Dear aspirant,

University College of Medical Sciences, New Delhi is a Public/ Government University which ranked 5 th in Medical Colleges in India. It is a accredited University and one of the reputed University. This College is affiliated to University of Delhi, New Delhi.

It offers both Undergraduation and Post

Hi aspirants

yes, Diploma holder can do admission in Bsc or BA in second year through lateral entry in distance mode. You do not have to take admission in 1 year you will be directed to 2 year of any undergraduate program.

Hope this Helps

Good Luck!

Hello Aspirant

hope you are doing good

With reference to your query, I would like to tell you that PGDM is a post graudate diploma management course in information technology which focus on technology manegment so that students can work efficently in a dyanmic IT environment . They have only

Hello Dear,

According to you, you got 5750 rank in TS EAMCET 2021. And you are  belonging to SC category then you have SC category quota. You are also non-local.There is good chance to get admission in JNTUH College Of Engineering, Hyderabad for CSE branch. But there is tough chance

Given

∑r=0^n2nCr

It can be written as

∑r=0nr⋅Cr2+∑r=0nCr2

We know that

r⋅nCr=n⋅n−1Cr−1

Hence

r⋅Cr2=n⋅n−1Cr−12

∑r=0nn⋅n−1Cr−12=n[n−1C02+n−1C12+.......+n−1Cn−12]=n⋅(n−1)!(2n−1−n+1)!(2n−1)!=n⋅(n−1)!n!(2n−1)!−−−−(1)

∑r=0nCr2=n!n!2n!−−−−−(2)

Adding both eq(1) and (2)

∑r=0n(r+1)Cr2=n⋅(n−1)!n!(2n−1)!+n!n!2n!

∑r=0n(r+1)Cr2=n⋅(n−1)!n!(2n−1)!+n(n−1)!n!2n(2n−1)!

∑r=0n(r+1)Cr2=(n−1)!n!(n+2)(2n−1)!

Ans:-   (n+2)(2n-1)! / n!(n-1)!