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Permutations and combinations are the processes of expressing an object selected from a set to form the required subsets.
- Permutations and Combinations select objects with or without replacement.
- The concept can be explained with the help of factorials.
- Permutations refer to the arrangement of elements in the set according to some order.
- Setting a sequence for a lock is an example of permutations.
- Combinations refer to the selection of objects in the set irrespective of any order.
- Selecting numbers for a lottery is an example of a combination.
- Terms like k-selection or k-combination are used for combinations with repetitions.
- This proves that permutations are known as arrangements, and combinations are known as selection.
- The formula used for permutations and combinations is as follows:
P(n,r) = n!/(n-r)!, where [n>= r]
C(n,r) = n!/[r !(n-r)!], where [n>= r]
- Where P(n,r) is permutation of elements
- C(n,r) is combination of elements
From an examination point of view, students can practice Important Questions For Class 11 Maths Chapter 7: Permutations and Combinations and NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combination.
Permutations and Combinations MCQs
Ques. Solve the value of: 5! – 2!.
- 118
- 119
- 121
- 112
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Ans. (a) 118
Explanation: 5! = 1×2×3×4x5 = 120
- 2! = 1 × 2 = 2
- Therefore, 5! – 2! = 120 – 2= 118
Ques. There are 20 chair patterns and ten table layouts available to a party planner. How many different ways can she create a set of tables and chairs for the party.
- 230
- 300
- 200
- 400
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Ans. (c)200
Explanation: The party planner has 20 chair designs and 10 table styles.
- There are 20 different ways to choose a chair.
- There are 10 different ways to choose a table.
- As a result, one chair and one table may be chosen in 20×10 = 200 different ways
Ques. How many squares can be created on a chessboard if there are 6 horizontal lines and 6 vertical lines.
- 110
- 245
- 225
- 125
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Ans. (c)225
Explanation: Total number of square = 6C2 × 6C2
- 15 × 15
- 225
Ques. Find the number of permutations if n = 12 and r = 2.
- 110
- 230
- 300
- 400
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Ans. (a) 110
Explanation: Given, n = 11 and r = 2
- Using the formula given above:
- nPr = (n!) / (n-r)! =(11!) / (11-2)! = 11! / 9! = (11 x 10 x 9! )/ 9! = 110
Ques. How many different ways can a team of 4 people be established from a total of 9 people so that two specific people are included in each team.
- 11
- 21
- 22
- 31
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Ans. (b) 21
Explanation: Each team should have two particular individuals on it. As a result, we must choose the remaining (4 – 2) = 2 people from a total of (9 – 2) = 7 people.
- As a result, the necessary number of ways has been met.
- 7C2
- (7×6)/(2×1)
- 7 x 3
- 21
Ques. How many 5-digit numbers can be formed with the digits 1 to 8 with no repetition allowed.
- 20
- 19
- 12
- 56
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Ans. (d) 56
Explanation: To fill nine digits in the place of a five-digit number, the order must be relevant.
- Here, the 5-digit numbers can be as many as there are permutations of 8 digits picked 5 at a time.
- Therefore, the five-digit number will be 8! / 5! x 3!
- 56
Ques. How many words can be created by using 3 letters from the term“LOVE”.
- 12
- 4
- 2
- 3
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Ans. (b) 4
Explanation: The word LOVE has 4 distinct letters.
- Therefore, the required number of words = 4P3 = 4! / (4 – 3)!
- Required number of words = 4! / 3! = 4
- Required number of words = 4
Ques. Determine the different combinations if you have 5 items and choose 3.
- 10
- 28
- 20
- 30
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Ans. (a) 10
Explanation: C(n, r) = n! / r! (n – r)!
- nCr = 5! / 3! (5 – 3)!
- nCr = (5 × 4 × 3 × 2 × 1) / ( 3 × 2 × 1 ) (2 x 1)
- nCr = 10
Ques. A pizza restaurant offers 3 different toppings for their pizzas. If a customer wants to order a pizza with exactly 2 toppings, in how many ways can this be done.
- 1
- 2
- 3
- 4
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Ans. (c)3
Explanation: It is a type of combination problem
- Now use the combination formula, we get:
- 3C2 = 3! / (2! x (3 – 2)!)
- 3! / (2! x 1!)
- (3) / (1 x 1)
- 3
Ques. How many 4-letter words can be formed using the letters from the word FABLE..
- 120
- 200
- 220
- 112
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Ans. (a) 120
Explanation: It is a type of permutation problem.
- Now use the permutation formula, we get:
- 5P4 = 5! / (5 – 4)! = 5! / 1! = 5 x 4 x 3 x 2 x 1= 120
Ques. Solve the value of: 7! – 6!.
- 4320
- 1190
- 1200
- 1100
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Ans. (a) 4320
Explanation: 7! = 1 × 2 × 3 × 4 x 5 x 6 x 7 = 5040
- 6! = 1 × 2 × 3 × 4 x 5 x 6 = 720
- Therefore, 7! – 6! = 4320
Ques. There are 30 chair patterns and eight table layouts available to a party planner. How many different ways can she create a set of tables and chairs for the party.
- 240
- 300
- 290
- 400
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Ans. (a) 240
Explanation: The party planner has 30 chair designs and 8 table styles.
- There are 30 different ways to choose a chair.
- There are 8 different ways to choose a table.
- As a result, one chair and one table may be chosen in 30×8 = 240 different ways
Ques. How many squares can be created on a chessboard if there are 4 horizontal lines and 6 vertical lines.
- 110
- 24
- 90
- 125
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Ans. (c)90
Explanation: Total number of square = 4C2 × 6C2
- 6 × 15
- 90
Ques. Find the number of permutations if n = 10 and r = 7.
- 10! / 3!
- 9! / 3!
- 8! / 3!
- 10! / 2!
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Ans. (a) 10! / 3!
Explanation: Given, n = 10 and r = 7
- Using the formula given above:
- nPr = (n!) / (n-r)! =(10!) / (10-7)! = 10! / 3!
Ques. How many different ways can a team of 6 people be established from a total of 9 people so that two specific people are included in each team.
- 11
- 35
- 20
- 31
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Ans. (b) 35
Explanation: Each team should have two particular individuals on it. As a result, we must choose the remaining (6 – 2) = 4 people from a total of (9 – 2) = 7 people.
- As a result, the necessary number of ways has been met.
- 7C4
- 35
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