## Vineet Loomba is teaching live on Unacademy Plus

PERMUTATIONS AND COMBINATIONS JEE MAIN AND ADVANCED (IIT-JEE) IIT-IEE MATHS MADE EASY PREPARED BY: ER. VINEET LOOMBA IITiAN | IIT-JEE MENTOR FoR SURE SHoT SUcCESS IN JEE MAIN AND ADVANCED (IIT-JEE)

REVISION COURSE FOR JEE MAIN AND ADVANCED (IIT-JEE) ABOUT ME B.Tech. From IIT Roorkee IIT-JEE Mentor Since 2010 Doubts/Feedback in Comment Section * Comment other topics you want to revise. Follow me @ https://unacademy.com/user/vineetloomba to get updates or search me on Google * Share among your peers as SHARING is CARING!!

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Other Detailed Courses Made so far on Unacademv: V Strategy for JEE Main and Advanced V Sets, Relations and Functions V Trigonometry V Applications of Derivatives Limits, Continuity. Differentiability V Indefinite Integration V Definite Integration Complex Number V Logarithmic Functions V Sequences Series V Most Important Questions in IIT-JEE Permutations Combinations Binomial Theorem V Straight Lines V Applications of Integrals V Circles (Detailed Course) V Probability MathematicS V Parabola (Detailed Course) v Inverse Trigonometry V Mathematical Reasoning V Ellipse (Detailed Course) Upcoming Courses next month: Coordinate Geometry > Differential Equations MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Ex. If Pr +654 P +3 30800 1 then the value of r is (A) 14 (B) 41 (C) 51 (D) 10 56630800 Sol. 54p1 56 ! (30800) x54! (56-r-6)! (54-r-3)! 56 55 (51-r) = 30800 30800 (51-56 x55 = 10 r 51-10 = 41 MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Ex. How many different words beginning with A and ending with L can be formed by using the letters of the word 'ANILMANGAL'? (A) 10080 (C) 20160 (B) 40320 (D) None of these Sol After fixing the letters A and L in the first and last places, the total number of available places are 8 and the letters are also 8. Out of these 8 letters there are 2 groups of alike letters. Therefore no. of words10080 8! 212! MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Ex. The number of different words (meaningful or meaningless) can be formed by taking four different letters from English alphabets is- (A) (26)4 (C) (25)4 The first letter of four letter word can be chosen by 26 ways, second by 25 ways, third by 24 ways and fourth by 23 ways. So number of four letter words 26 x 25 x 24 x 23 358800 (B) 358800 (D) 15600 Sol. MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) 52-r Ex. 47C + 4 3 is equal to- (A) 51CA (C) 53C4 (B) 52C4 (D) None of these Sol.The given expression can be written as 52-r C3 + 47c 4 47c + - 51C3 +50C3 + 49C3 +48C3 + 41C34C4 [We know that C+1CnC 550C3 49C348C48C - 5'C3 + 50c3 + 49C34C4 5150C50C 51C+51C4 -52C4 MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) A candidate is required to answer 6 out of 10 questions which are divided into two groups each containing 5 questions and he is not permitted to attempt more than 4 from each group. The number of ways in which he can make up his choice is- (A) 100 (B) 200 (C) 300 (D) 400 Ex. So Let there be two groups A and B each containing 5 questions. Questions to be attempted is 6, but not more than 4 from any group. The candidate can select the questions in following ways: (i) 4 from group A and 2 from group B. () 3 from group A and 3 from group B. (ii) 2 from group A and 4 from group EB The number of selections in the above cases are 5C4 x 5C2,503 5C3,5C2 x 5C4 respectively. :. Number of ways of selecting 6 questions -5C4* 5C2+ 5C3* 5C3+ 5C2 x 5C4 -5010050 200 MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) There are four balls of different colours and four boxes of colours same as those of the balls. The number of ways in which the balls, one each box, could be placed such that a ball does not go to box of its own colour is- (A) 8 (C) 9 Number of derangements are Ex. (B) 7 (D) None of these Sol. 12 41 9 (Since number of derangements in such a problems is given by n! 1! 2 3-1)" MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) ENROL for this Course (Free) Enroll 439 Recommend Lessons (Like) E . Rate and Review the Course .Comments . Sharing with friends 15 1. 68 26 ratings 7 reviews Share MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)