## Vineet Loomba is teaching live on Unacademy Plus

EE MAIN AND ADVANCED RANK BoOSTER COURSE PREVIOUS YEAR QUESTIONS PERMUTATIONS COMBINATIONS JEE MAIN 2019 PREPARED BY: ER. VINEET LOOMBA IITIAN IIT-IEE MENTOR

SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) ABOUT ME B. Tech. From IIT Roorkee IIT-JEE Mentor Since 2010. .,. Verified Star Educator @ Unacademy Youtuber. Founder @ vineetloomba.com unacademy Many of my students are pursuing courses in IITs, NITs etc. Search "vineet loomba unacademy on GOOGLE

EE MAIN AND ADVANCED RANK BoOSTER COURSE PREVIOUS YEAR QUESTIONS PERMUTATIONS COMBINATIONS JEE MAIN 2019 PREPARED BY: ER. VINEET LOOMBA IITIAN IIT-IEE MENTOR

SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Other Detailed Courses Made so far on Unacademy: Strategy for JEE Main and Advanced Sets, Relations and Functions Trigonometry Applications of Derivatives Permutations Combinations Binomial Theorem Straight Lines Applications of Integrals Circles (Detailed Course) Probability Limits, Continuity, Differentiability Indefinite Integration ..#finite Into(Jtation Complex Number Logarithmic Functions Sequences Series Most Important Questions in IIT-JEE Mathematics Parabola (Detailed Course) Inverse Trigonometry Mathematical Reasoning Ellipse (Detailed Course) Hyperbola Differential Equations Upcoming Courses next month: Statistics Matrices Determinants MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT ROORKEE)

SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Example: Consider a class of 5 girls and 7 boys. The number of different teams consisting of 2 girls and 3 boys that can be formed from this class, if there are two specific boys A and B, who refuse to be the members of the same team, is (1) 200 (3) 500 (JEE Main 2019) (2) 350 (4) 300 Firstly select 2 girls by 5C2 ways. Solution: 3 boys can be selected in following 3 scenarios. (i) Selection of A and selection of any 2 other boys (except B) C2 ways (ii) Selection of B and selection of any 2 other boys (except A)52 ways (ii) Selection of 3 boys (except A and B) C3 ways Number of ways = 5C2 (5C2 + 5C2 + 5G)= 300 MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT ROORKEE)

SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) The sum of all two digit positive numbers which when divided by7 yield 2 or 5 as remainder is (1) 1465 (3) 1365 Example: (JEE Main 2019) (2) 1356 (4) 1256 Solution: S2 12 + 19 + 26+.... +96 702 Required Sum 654 702- 1356 MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT ROORKEE)

SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Example: The number of functions f from 11, 2, 3, .20 onteo 11, 2, 3, ..., 20} such that f(k) is a multiple of 3, whenever k is a multiple of 4, is: (1) 56 15 (3) 5! x 6! (2) 65 x (15)! JEE Main 2019) (4) (15)! x 6! Domain and codomain 1, 2, 3,, 20 Multiples of 4 4, 8, 12, 16 and 20 Multiples of 3 3, 6, 9, 18. 15. 18. If k is multiple of 4 then f(k) is multiple of 3 and function is onto Total number of arrangement= 6C5 5! = 6! Remaining 15 can arrange in 15! ways. Total number of arrangement = 1516! MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT ROORKEE)

SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Example: There are m men and two women participating in a chess tournament. Eaclh participant plays two games with every other participant. If the number of games played by the men between themselves exceeds the number of games played between the men and the women by 84, then the value of m is (2) 7 (4) 1 (JEE Main 2019) (3) 11 Solution: m(m-1) = 4m + 84 m2-12m-7m-84 = 0 m(m-12) +7 (m-12) = 0 : m> 0 m12 MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT ROORKEE)

SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) If nC4,ng, and nC6 are in . ., then n can be: (1) 12 (3) 14 Example: (2) 9 (4) 11 (JEE Main 2019) 12(n-4) = 30 + n2-9n + 20 n2 21n +98 0 (n-7) (n-14) = 0 n=7, n = 14 nCsC 5 n-5 n-4 6 MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT ROORKEE)

SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Example: Consider three boxes, each containing 10 balls labelled 1, 2, ....10. Suppose one ball is randomly drawn from each of the boxes. Denote by n, the label of the ball drawn from the ith box, (i = 1, 2, 3). Then, the number of ways in which the balls can be chosen such that ni < n2 < n3 is (1) 240 (3) 164 (JEE Main 2019) (2) 120 (4) 82 NOW of Different labels of balls drawn 10 x 9 x8 10 x 9 x 8 3! As arrangement is not required = 120 MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT ROORKEE)