JEE Main 27 August 2021, Shift 1, Maths Q80

Question 80 of 90 in this shift, Maths question 20 of 30, Section A.

\sum_{k=0}^{20}\left({}^{20}C_k\right)^2

is equal to :

(1) ${}^{41}C_{20}$
(2) ${}^{40}C_{19}$
(3) ${}^{40}C_{20}$ Official answer
(4) ${}^{40}C_{21}$

Official answer

Option 3

NTA final key.

Chapter: Binomial Theorem
Topic: Binomial Coefficients
Idea tested: Sum of Squares of Binomial Coefficients

Same idea in other shifts

Asked 2× in all

17 Mar 2021, Shift 2 · Q76The value of $\sum_{r=0}^{6}\left({}^6C_r\cdot{}^6C_{6-r}\right)$ is equal to :EasySingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.