Class 11 RD Sharma Solutions- Chapter 17 Combinations- Exercise 17.1 | Set 1 Last Updated : 23 Jul, 2025 Comments Improve Suggest changes Like Article Like Report Chapter 17 of RD Sharma’s Class 11 Mathematics textbook focuses on Combinations a crucial topic in combinatorics. Combinations involve selecting items from the larger set where the order of the selection does not matter. This chapter aims to develop an understanding of how to calculate and analyze different ways of choosing items that have practical applications in probability, statistics, and various problem-solving scenarios.CombinationsThe Combinations refer to the selection of the items from a set where the order does not matter. For a set of n items the number of ways to choose r items is given by the combination formula:\binom{n}{r} = \frac{n!}{r!(n-r)!} where n! denotes the factorial of n. This formula is fundamental for solving problems involving the selection and arrangement where the sequence of the choices is irrelevant.Question 1. Evaluate the following:i) 14C3Solution:We know that nCr=n!/(n-r)!r!=>14C3=14!/(14-3)!3! =14!/11!3! =14x13x12/3x2x1 =364ii) 12C10 Solution:= 12!/(12-10)!10!= 12!/2!10!= 12x11/2x1= 66iii) 35C35Solution:= 35!/(35-35)!35!= 1iv) n+1CnSolution:= (n+1)!/(n+1-n)!n!= (n+1)!/n!= n+1v) 5Solution:∑ 5Cr=5C1+5C2+5C3+5C4+5C5r = 1= 5+10+10+5+1= 31Question 2. If nC12=nC5, find the value of n.Solution:Given that nC12=nC5.We know that two combinations will be equal when the sum of their r's is equal to n.=>n=12+5=17.Question 3. If nC4=nC6 , find 12Cn.Solution:=>n=6+4=10=>12C10=12!/10!2! =12x11/2 =66Question 4. If nC10=nC12 , 23Cn.Solution:n = 10+12=22=>23C22 = 23!/22!1! = 23Question 5. If 24Cx=24C2x+3 , find x.Solution:24 = x+2x+324 = 3x+321 = 3xx = 21/3x = 7Question 6. If 18Cx=18Cx+2 , find x.Solution: 18 = x+x+218 = 2x+216 = 2xx = 8Question 7. If 15C3r=15Cr+3, find r.Solution: 15 = 3r+r+315 = 4r+312 = 4rr = 3Question 8. If 8Cr-7C3=7C2, find r.Solution: Given 8Cr-7C3=7C2=>8Cr=7C2+7C3We know that nCr+nCr-1=n+1Cr=>8Cr=8C3=>r=3Question 9. If 15Cr:15Cr-1 = 11:5, find r.Solution: 15Cr/15Cr-1=11/5(15!/(15-r)!r!)/(15!/(15-r+1)!(r-1)!)=11/515-r+1/r = 11/55(16-r) = 11r80-5r = 11r16r = 80r = 5Question 10. If n+2C8:n-2P4=57:16, find n.Solution:We know that nPr=n!/(n-r)!=>((n+2)!/(n+2-8)!8!)/((n-2)!/(n-2-4)!)=57/16=>(n+2)(n+1)(n)(n-1)/8!=57/16=>(n-1)n(n+1)(n+2)=(57/16)8!=>(n-1)n(n+1)(n+2)=57x7!/2=>(n-1)n(n+1)(n+2)=57x7x6x5x4x3=>(n-1)n(n+1)(n+2)=19x3x7x6x5x4x3=>(n-1)n(n+1)(n+2)=19x18x20x21=>n=19Read More:CombinationsCombinations Formula with ExamplesConclusionUnderstanding combinations is essential for the solving problems in probability and statistics where the order of the selection is not a factor. Mastery of the combination formula and its applications allows for the accurate calculation of possible outcomes in various scenarios enhancing problem-solving skills in the mathematics. Comment More infoAdvertise with us T thanmaig142 Follow Improve Article Tags : Mathematics School Learning Class 11 RD Sharma Solutions RD Sharma Class-11 Maths-Class-11 +2 More Similar Reads Maths Mathematics, often referred to as "math" for short. It is the study of numbers, quantities, shapes, structures, patterns, and relationships. It is a fundamental subject that explores the logical reasoning and systematic approach to solving problems. Mathematics is used extensively in various fields 5 min read Basic ArithmeticWhat are Numbers?Numbers are symbols we use to count, measure, and describe things. 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