NCERT Solutions for Class 11 Maths Exercise 7.4 of the Chapter 7 Permutaion and Combination
NCERT Solutions for Class 11 Maths Exercise 7.4 of the Chapter 7 Permutaion and Combination are available here for boosting the preparation of class 11 maths exam. The NCERT Solutions for Class 11 Maths Exercise 7.4 of the Chapter 7 Permutaion and Combination are created by an expert of maths who has huge experience in teaching maths from class 9 to class 12.All the questions of permutation and combination are solved by step by step ways therefore every students of the grade 11 can understand the solutions easily .
NCERT Solutions for Class 11 Maths Chapter 7 Permutation and Combination
Exercise 7.1- Permutations and Combinations
Exercise 7.2 -Permutation and Combination
Exercise 7.3-Permutation and Combination
Exercise 7.4-Permutation and Combination
Miscellaneous Exercise-Permutation and Combination
NCERT Solutions for Class 11 Maths Exercise 7.4 of the Chapter 7 Permutation and Combination
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Q1. If ^{n}C_{8} = ^{n}c_{2}, find ^{n}c_{2}
Ans. Since ^{n}C_{r }=^{n}C_{n-r}
∴^{n}C_{8 }= ^{n}C_{n-8, }
Substituting ^{n}C_{8 }= ^{n}C_{n-8,}
^{n}C_{n-8} = ^{n}c_{2}
n – 8 = 2⇒ n = 10
Hence the required value of ^{n}c_{2}is 45
Q2.Determine n if
(i) ^{2n}C_{3}: ^{n}c_{3 }= 12 : 1
(ii) ^{2n}C_{3}: ^{n}c_{3 }= 11 : 1
Ans. It is given to us that
^{2n}C_{3}: ^{n}c_{3 }= 12 : 1
4(2n -1)(n -1) : (n -1)(n -2) = 12 : 1
8n -4 = 12n – 24
4n = 20
n = 5
(ii) ^{2n}C_{3}: ^{n}c_{3 }= 11 : 1
Similarly as in (i),we have
11n – 22 = 8n – 4
3n = 18
n = 6
Q3. How many chords can be drawn through 21 points on a circle?
Ans.For drawing chords we are required to choose two points out of 21 points on the circle, therefore the number of ways of such combinations are ^{21}C_{2}
The number of chords are = ^{21}C_{2}
Hence total number of chords can be drawn are 210
Q4.In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls.
Ans. The number of ways 3 boys can be selected out of 5 boys are ^{5}C_{3}
The number of ways 3 girls can be selected out of 4 girls are ^{4}C_{3}
According to the multiplication rule,total number of ways selecting a team of 3 boys and 3 girls are = ^{5}C_{3 }×^{4}C_{3}
Hence the number of ways a team of 3 boys and 3 girls selected from 5 boys and 4 girls is 40
Q5.Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.
Ans. The number of ways of selecting 3 balls out of 6 red balls are ^{6}C_{3}
The number of ways of selecting 3 balls out of 5 white balls are ^{5}C_{3}
The number of ways of selecting 3 balls out of 5 blue balls are ^{5}C_{3}
According to the multiplication rule,total number of ways selecting 9 balls out of 6 red ball,5 white balls and 5 blue balls are = ^{6}C_{3 }×^{5}C_{3 }×^{5}C_{3}
= 20 × 10 × 10 = 2000
Hence required number of ways of selecting 9 balls are 2000
Q6.Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.
Ans. The number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination are =The number of ways selecting 1 ace out of 5 cards × The number of ways selecting remaining 4 cards out of remaining (52 -4= 48 ) cards
The number of ways selecting 1 ace out of 4 cards = ^{4}C_{1}
The number of ways selecting remaining 4 cards out of remaining (52 -4 = 48) cards = ^{48}C_{4}
The number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination are = ^{4}C_{1 }× ^{48}C_{4}
= 4× 2× 47 ×46 ×45 = 778320
The number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination is 778320
Q7.In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers.
Ans.The number of ways selecting a team of 11 players = The number of ways selecting 4 bowlers out of 5 bowlers × The number of ways selecting 7 players out of remaining (17-5 =12) players
The number of ways selecting a team of 11 players = ^{5}C_{4}×^{12}C_{7}
=5 × 11×9×8 = 3960
The number of ways selecting a team of 11 players in which there are exactly 4 bowlers is 3960
Q8.A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.
Ans.The number of ways selecting 2 black balls out of 5 black balls = ^{5}C_{2}
The number of ways selecting 3 red balls out of 6 red balls = ^{6}C_{3}
The number of ways selecting 2 black balls and 3 red balls out of 5 black balls and 6 red balls =^{5}C_{2 }× ^{6}C_{3}
= 10 × 20 = 200
The number of ways selecting 2 black balls and 3 red balls out of 5 black balls and 6 red balls is 200
Q9.In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?
Ans. It is given that student has to choose 5 coarses out of 9 coarses in which 2 coarses are compulsury.
Since 2 coarses are compusury therefore number of ways selecting 5 coarses = Number of ways of selecting remaining (5 -2 = 3) coarses out of remaining (9 -2 = 7) coarses is = ^{7}C_{3}
Hence required number of ways selecting 5 coarses is 35
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Chapter 2- Relations and functions | Chapter 10- Straight Lines |
Chapter 3- Trigonometry | Chapter 11-Conic Sections |
Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |
Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |
Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |
Chapter 7- Permutations and Combinations | Chapter 15- Statistics |
Chapter 8- Binomial Theorem | Chapter 16- Probability |
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Chapter 1-Relations and Functions | Chapter 9-Differential Equations |
Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |
Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |
Chapter 4-Determinants | Chapter 12-Linear Programming |
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