## NCERT Solutions Class 11th Maths 7: Permutations And Combinations## Exercise 7.1
**repetition of the digits is allowed?****repetition of the digits is not allowed?**
Let A be the hundredths place digit, B be the tens place digit, and C be the units place digit of a 3-digit number. (i) Since, repetition is allowed. Therefore, A, B, and C can all have any of the 5 digits available. Hence, total number of 3-digit numbers that can be formed = 5 × 5 × 5 = 125 (i) Since, repetition is not allowed. Therefore, if C occupies one of the 5 available digits then B has to be one of the 4 available digits. Similarly, when B occupies one of the remaining 4 digits then A has to be one of the 3 available digits. Hence, total number of 3-digit numbers that can be formed = 5 × 4 × 3 = 60
Let A be the hundredths place digit, B be the tens place digit, and C be the units place digit of a 3-digit number. In order for the number ABC to be even, the units place digit has to be even. This implies that C has to be one of either 2, 4 or 6. So, C can be one of 3 possible digits. Since, repetition is allowed. Therefore, both A and B can have any one of the 6 given digits. Hence, total number of 3-digit even numbers that can be formed = 6 × 6 × 3 = 108
Let there be a 4-letter code αβγδ. Since, repetition is not allowed. Therefore, if α occupies one of the 10 available alphabets, then β has to be one of the 9 remaining ones. Similarly, when β occupies one of the available 9 alphabets then γ has to be one of the 8 remaining ones and when γ occupies one of the 8 available alphabets then δ has to be one of the 7 remaining ones. Hence, total number of 4-letter codes possible = 10 × 9 × 8 × 7 = 5040
Let the 5-digit telephone number be ABCDE where AB is fixed to be 67, C is hundredths place digit, D is tens place digit, and E is units place digit. 6 and 7 are already used once and cannot appear again, which leaves 8 digits for the rest of the number. Since, repetition is not allowed. Therefore, if C occupies one of the 8 available digits then D has to be one of the 7 remaining ones. Similarly, when D has to be one of the available 7 digits then E has to be one of the 6 remaining ones. Hence, total number of 5-digit telephone numbers that can be constructed = 1 × 1 × 8 × 7 × 6 = 336
When a coin is tossed, it has 2 possible outcomes - either heads or tails. When, a coin is tossed 3 times, it is possible for it to show one of the two outcomes on each toss. Hence, total possible outcomes = 2 × 2 × 2 = 8
The upper flag can have any of the 5 available colours. Since, repetition is not allowed. Therefore, the lower flag can use any one of the remaining 4 colours. Hence, total possible number of signals that can be generated = 5 × 4 = 20 ## Exercise 7.2
= 40320
4! = 4 × 3 × 2 × 1 which can be written as 4 × 3! Therefore, 4! - 3! = 4 × 3! - 3! = 3! × (4 - 1) = 3 × 2 × 1 × (3) = 18
LHS = 3! + 4! = 3! + 4 × 3! = 3! × (4 + 1) = 5 × 3! RHS = 7! = 7 × 6 × 5 × 4 × 3! = 7 × 6 × 4 × (5 × 3!) Clearly, 7! > 3! + 4!. LHS ≠ RHS Hence, 3! + 4! ≠ 7!
8!/(6! × 2!) = (8 × 7 × 6!)/(6! × 2 × 1) = (8 × 7)/2 = 4 × 7 = 28
LHS = 1/6! + 1/7! = 1/6! + 1/(7 × 6!) = 1/6! × (1 + 1/7) = 1/6! × (7 + 1)/7 = 1/6! × 8/7 = 8/(7 × 6!) = 8/7! Now, 8/7! = x/8! 8/7! = x/(8 × 7!) 8 = x/8 x = 64
n!/(n - r)! = 6!/(6 - 2)! = 6!/4! = (6 × 5 × 4!)/4! = 6 × 5 = 30
n!/(n - r)! = 9!/(9 - 5)! = 9!/4! = (9 × 8 × 7 × 6 × 5 × 4!)/4! = 9 × 8 × 7 × 6 × 5 = 15120 ## Exercise 7.3
Total number of possible digits = 9 = n The number needs to have 3 digits. So, r = 3 It is given that repetition is not allowed. Therefore, Number of 3-digit numbers from 1 to 9 without repetition = Possible number of permutations =
= 9 × 8 × 7 × 6!/6! = 504
The number at thousands place cannot be 0, otherwise it would become a 3-digit number. Thus, thousands place digit can only have 9 possible digits from 1 to 9. Number of possible digits for the remaining 3 digits of the number = 9 = n We need to select 3 digits from the remaining 9. So, r = 3 It is given that repetition is not allowed. Therefore, Number of 4-digit numbers without repetition = 9 × Possible number of permutations = 9 × 9 × = 9 × 9 × 8 × 7 × 6!/6! = 4536
In order for the number to be even, the units place digit has to be either 2, 4 or 6. As repetition is not allowed, number of digits left = 5 The remaining two digits of the number can be formed in
= 5 × 4 × 3!/3! = 20 Total possible 3-digit even numbers with given digits = 3 ×
Total number of available digits = 5 Number of 4-digit numbers without repetition = = 5 × 4 × 3 × 2 × 1/1! = 120 In order for the number to be even, the units place digit has to be either 2 or 4. As repetition is not allowed, number of digits left = 4 The remaining 3 digits of the number can be formed in
= 4 × 3 × 2 × 1/1! = 24 Number of even 4-digit numbers without repetition = 2 ×
Total possible candidates for chairman and vice chairman = n = 8 Number of people to be selected = r = 2 Total number of ways to choose a chairman and a vice chairman = Number of Permutations
= 8 × 7 × 6!/6! = 56
= (n - 1) × (n - 2) × (n - 3) × (n - 4)!/(n - 4)! = (n - 1)(n - 2)(n - 3)
= n × (n - 1) × (n - 2) × (n - 3) × (n - 4)!/(n - 4)! = n(n - 1)(n - 2) Now, (n - 1)(n - 2)(n - 3)/n(n - 1)(n - 2)(n - 3) = 1/9 1/n = 1/9 n = 9
5!/(5 - r)! = 2 × (6!/(6 - r + 1)!) 5!/-(r - 5)! = 2 × (6!/-(r - 7)!) 5!/(r - 5)! = 2 × 6!/(r - 7)! (r - 7)!/(r - 5)! = 2 × 6!/5! (r - 7) × (r - 6) × (r - 5)!/(r - 5)! = 2 × 6 × 5!/5! (r - 7)(r - 6) = 12 r r r r(r - 10) - 3(r - 10) = 0 (r - 10)(r - 3) = 0 When, (r - 10) = 0, r = 10 OR When, (r - 3) = 0, r = 3 But, r = 10 is not possible as it is greater than n = 5 in Hence, r = 3.
5!/(5 - r)! = (6!/(6 - r + 1)!) 5!/-(r - 5)! = (6!/-(r - 7)!) 5!/(r - 5)! = 6!/(r - 7)! (r - 7)!/(r - 5)! = 6!/5! (r - 7) × (r - 6) × (r - 5)!/(r - 5)! = 6 × 5!/5! (r - 7)(r - 6) = 6 r r r r(r - 9) - 4(r - 9) = 0 (r - 9)(r - 4) = 0 When, (r - 9) = 0, r = 9 OR When, (r - 4) = 0, r = 4 But, r = 9 is not possible as it is greater than n = 5 in Hence, r = 4.
Unique letters in the word EQUATION = {E, Q, U, A, T, I, O, N} = 8 letters Number of letters to be used for making a word = 8 Total number of possible words = Total number of permutations
**4 letters are used at a time,****all letters are used at a time,****all letters are used but first letter is a vowel?**
Unique letters in the word MONDAY = {M, O, N, D, A, Y} = 6 letters
Possible number of permutations = = 6!/2! = 720/2 = 360
Possible number of permutations = = 6!/0! = 720
Number of letters left to be used = 5 Number of letters to be used for making the word = 5 Possible number of permutations = = 5!/0! = 120 Possible number of words = 2 × 180 = 240
Total number of letters in MISSISSIPPI = 11 Unique letters = {M, I, S, P} M occurs once, I occurs four times, S occurs four times, and P occurs two times. Number of distinct permutations = 11!/(1!4!4!2!) = 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4!/(4! × 4! × 2) = 11 × 10 × 9 × 8 × 7 × 6 × 5/(48) = 34650 If the four I's are always together they can treated as 1 letter whose position affects the permutations. Thus, remaining number of letters = 11 - 4 + 1 = 8 Number of permutations where fours I's are together = 8!/1!4!2! = 8 × 7 × 6 × 5 × 4!/(4! × 2) = 840 Hence, total number of permutations where four I's are not together = 34650 - 840 = 33810
**words start with P and end with S,****vowels are all together,****there are always 4 letters between P and S?**
Unique letters in the word PERMUTATIONS = {P, E, R, M, U, T, A, I, O, N, S} = 11 letters
If P and S are fixed as the first and last letters then remaining number of letters = 12 - 2 = 10 Number of permutations for the remaining 10 letters = Total number of permutations when P and S are fixed = = 10!/2(10 - 10)! = 10!/2(0!) = 1814400
If the five vowels are always together they can treated as 1 letter whose position affects the permutations. Thus, remaining number of letters = 12 - 5 + 1 = 8 Number of permutations = = 8!/2(8 - 8)! = 8!/2(0!) = 20160
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 Since, there needs to be 4 letters between P and S. Therefore, possible positions of P and S can be 1 and 6, 2 and 7, 3 and 8, 4 and 9, 5 and 10, 6 and 11, 7 and 12 respectively. Thus, there are 7 possible ways to place P and S such that there are 4 letters between them. P and S can also be interchanged in these positions, so total possible ways to arrange P and S = 2 × 7 = 14 Remaining letters = 12 - 2 = 10 Number of permutations for the remaining letters = = 10!/2(10 - 10)! = 10!/2(0!) = 1814400 Total number of permutations = 1814400 × 14 = 25401600 ## Exercise 7.4
We know that when
So, 8 = n - 2 n = 10 Therefore,
= 10!/2(8!) = 10 × 9 × 8!/(2 × 8!) = 45
2n!/3!(2n - 3)! = 12 × n!/3!(n - 3)! 2n × (2n - 1) × (2n - 2) × (2n - 3)!/3!(2n - 3)! = 12 × n × (n - 1) × (n - 2) × (n - 3)!/3!(n - 3)! 2n × (2n - 1) × 2(n - 1) = 12 × n × (n - 1) × (n - 2) 2(2n - 1) = 6 × (n - 2) 4n - 2 = 6n - 12 0 = 2n - 10 2n = 10 n = 5
2n!/3!(2n - 3)! = 11 × n!/3!(n - 3)! 2n × (2n - 1) × (2n - 2) × (2n - 3)!/3!(2n - 3)! = 11 × n × (n - 1) × (n - 2) × (n - 3)!/3!(n - 3)! 2n × (2n - 1) × 2(n - 1) = 11 × n × (n - 1) × (n - 2) 4(2n - 1) = 11 × (n - 2) 8n - 4 = 11n - 22 0 = 3n - 18 3n = 18 n = 6
We know that a chord can be drawn by connecting 2 points on a circle. Therefore, Number of chords that can be drawn through 21 points = = 21!/(2 × 19!) = 21 × 20 × 19!/(2 × 19!) = 210 Total number of chords that can be drawn is 210.
Number of boys = 5 Number of boys to be selected = 3 Number of ways 3 boys can be selected from 5 boys = Number of girls = 4 Number of girls to be selected = 3 Number of ways 3 girls can be selected from 4 girls = Total number of ways to select the team = = 5!/3!(5 - 3)! × 4!/3!(4 - 3)! = 5!/3!2! × 4!/3!1! = 5 × 4 × 3!/(3! × 2) × 4 × 3!/3! = 10 × 4 = 40 Hence, total number of ways to select a team of 3 boys and 3 girls from 5 boys and 4 girls = 40
Balls to be selected from each colour = 3 Number of ways to select 3 red balls from 6 red balls = Number of ways to select 3 white balls from 5 white balls = Number of ways to select 3 blue balls from 5 blue balls = Total number of ways to select the balls = = 6!/3!(6 - 3)! × 5!/3!(5 - 3)! × 5!/3!(5 - 3)! = 6!/3!3! × 5!/3!2! × 5!/3!2! = 6 × 5 × 4 × 3!/3!3! × 5 × 4 × 3!/(3! × 2) × 5 × 4 × 3!/(3! × 2) = 6 × 5 × 4/(3 × 2) × 5 × 4/2 × 5 × 4/2 = 20 × 10 × 10 = 2000 Hence, total number of ways to select 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour = 2000
We know that in a deck of 52 cards, there are 4 ace cards. Since, we need to have one ace in each combination therefore, number of ways to select 1 ace from 4 ace cards = Remaining cards in the deck = 52 - 4 = 48 Remaining cards to be selected = 5 - 1 = 4 Number of ways to select 4 cards from 48 cards = Total number of ways to select the given card combinations = = 4!/1!(4 - 1)! × 48!/4!(48 - 4)! = 4!/3! × 48!/4!44! = 4 × 3!/3! × 48 × 47 × 46 × 45 × 44!/4!44! = 4 × 48 × 47 × 46 × 45/24 = 778320 Hence, total number of ways to select 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination = 778320
5 players in the team can bowl and we need to select 4 bowlers for the team of 11. Number of ways to select 4 bowlers from 5 bowlers = Remaining players = 17 - 5 = 12 Remaining spots in the team = 11 - 4 = 7 Number of ways to select 7 players from 12 players = Total number of ways to select the team = = 5!/4!(5 - 4)! × 12!/7!(12 - 7)! = 5!/4!1! × 12!/7!5! = 1/4! × 12!/7! = 1/24 × 12 × 11 × 10 × 9 × 8 × 7!/7! = 1/24 × 12 × 11 × 10 × 9 × 8 = 3960 Hence, total number of ways to select a cricket team of eleven from 17 players in which exactly 4 players are bowlers = 3960
Number of black balls = 5 Number of black balls to be selected = 2 Number of ways 2 blacks balls can be selected from 5 black balls = Number of red balls = 6 Number of red balls to be selected = 3 Number of ways 3 red balls can be selected from 6 red balls = Total number of ways to select the balls = = 5!/2!(5 - 2)! × 6!/3!(6 - 3)! = 5!/2!3! × 6!/3!3! = 5 × 4 × 3!/(2 × 3!) × 6 × 5 × 4 × 3!/3!3! = 5 × 4/2 × 6 × 5 × 4/(3 × 2) = 200 Hence, total number of ways to select 2 black and 3 red balls from 5 black balls and 6 red balls = 200
Total available courses = 9 2 of the courses are compulsory for every student. So, Remaining available courses = 9 - 2 = 7 Remaining courses to be selected = 5 - 2 = 3 Number of ways to select 3 courses from 7 courses = = 7!/3!4! = 7 × 6 × 5 × 4!/3!4! = 7 × 6 × 5/(3 × 2) = 35 Hence, total number of ways to select 5 courses from 9 courses when 2 specific courses are compulsory = 35 ## Miscellaneous Exercise
Vowels in the word DAUGHTER = {A, U, E} = 3 Consonants in the word DAUGHTER = {D, G, H, T, R} = 5 Number of vowels to be chosen = 2 Number of ways to select 2 vowels from 3 vowels = = 3!/2!1! = 3 × 2!/2! = 3 Number of consonants to be chosen = 3 Number of ways to select 3 consonants from 5 consonants = Number of ways to choose the letters for the word = = 3!/2!(3 - 2)! × 5!/3!(5 - 3)! = 3!/2!1! × 5!/3!2! = 3 × 2!/2! × 5 × 4 × 3!/3!2! = 3 × 10 = 30 Therefore, we have 2 vowels + 3 consonants = 5 letters to arrange and form words. Number of permutations = Hence, total number of words with 2 vowels and 3 consonants that can be formed from the word DAUGHTER = 30 × 120 = 3600
Vowels in the word EQUATION = {E, U, A, I, O} = 5 Consonants in the word EQUATION = {Q, T, N} = 3 Number of permutations for arranging the 5 vowels = = 5!/(5 - 5)! = 5!/0! = 120 Number of permutations for arranging the 3 consonants = = 3!/(3 - 3)! = 3!/0! = 6 When vowels and consonants appear together, either vowels will appear first or consonants will appear first, so we have two ways to arrange them. Hence, total number of words that can be formed from the word EQUATION when vowels and consonants are together = 2 × 6 × 120 = 1440
= 4!/3!1! = 4 × 3!/3! = 4 Remaining number of committee members to be selected = 7 - 3 = 4 Number of ways to select 4 boys from 9 boys = = 9!/4!5! = 9 × 8 × 7 × 6 × 5!/4!5! = 9 × 8 × 7 × 6/(4 × 3 × 2) = 126 Total number of ways to form a committee with exactly 3 girls = 4 × 126 = 504
When committee consists of 3 girls and 4 boys, it will be same as in When committee consists of 4 girls and 3 boys: Number of ways to select 4 girls from 4 girls is only 1. Number of ways to select 3 boys from 9 boys = = 9!/3!6! = 9 × 8 × 7 × 6!/3!6! = 9 × 8 × 7/(3 × 2) = 84 Total number of ways to form a committee with at least 3 girls = 84
0 girls, 7 boys 1 girl, 6 boys 2 girls, 5 boys 3 girls, 4 boys When committee consists of 3 girls and 4 boys, it will be same as in When committee consists of 0 girls and 7 boys: Number of ways to select 7 boys from 9 boys = = 9!/7!(9 - 7)! = 9!/7!2! = 9 × 8 × 7!/7!(2) = 36 Committee can be formed in When committee consists of 1 girl and 6 boys: Number of ways to select 1 girl from 4 girls = = 4!/3! = 4 × 3!/3! = 4 Number of ways to select 6 boys from 9 boys = = 9!/6!3! = 9 × 8 × 7 × 6!/3!6! = 9 × 8 × 7/(3 × 2) = 84 Committee can be formed in 4 × 84 = When committee consists of 2 girls and 5 boys: Number of ways to select 2 girl from 4 girls = = 4!/2!2! = 4 × 3 × 2!/(2 × 2!) = 6 Number of ways to select 5 boys from 9 boys = = 9!/5!4! = 9 × 8 × 7 × 6 × 5!/5!4! = 9 × 8 × 7 × 6/(4 × 3 × 2) = 126 Committee can be formed in 6 × 126 =
Words are arranged alphabetically in a dictionary, so we have to count the words that start with A, B, C or D. Out of these four, the word EXAMINATION only has A. So, we need to find the number of words that start with A. Total number of letters in the word = 11 Remaining letters available = 11 - 1 = 10 Number of permutations = 10!/2!2! = 10!/(2 × 2) = 907200 Hence, number of words in the dictionary before the word starting from E = 907200
We know that a number will only be divisible by 10 if the unit place has a 0. Therefore, in the 6-digit number to be formed, the units place will be fixed as 0. Remaining places to be filled in the number = 6 - 1 = 5 Number of permutations for the 5 digits = = 5!/(5 - 5)! = 5!/0! = 120 Hence, number of 6-digit numbers formed from 0, 1, 3, 5, 7 and 9 which are divisible by 10 = 120
Number of ways to choose 2 vowels from 5 vowels = Number of ways to choose 2 consonants from 21 consonants = Total number of ways to select the vowels and consonants = = 5!/2!(5 - 2)! × 21!/2!19! = 5!/2!3! × 21!/2(19!) = 5 × 4 × 3!/(2 × 3!) × 21 × 20 × 19!/2(19!) = 10 × 210 = 2100 The selected 4 letters can be arranged in different ways to make different words. Number of permutations = Total number of words that can be formed in the given manner = 24 × 2100 = 50400
The student can select: 3 questions from part I and 5 questions from part II, 4 questions from part I and 4 questions from part II, 5 questions from part I and 3 questions from part II When a student chooses 3 questions from part I and 5 questions from part II: Number of ways to select 3 questions from 5 = Number of ways to select 5 questions from 7 = Number of ways to select questions = = 5!/3!(5 - 3)! × 7!/5!(7 - 5)! = 5!/3!2! × 7!/5!2! = 5 × 4 × 3!/(2 × 3!) × 7 × 6 × 5!/(2 × 5!) = 10 × 21 = 210 When a student chooses 4 questions from part I and 4 questions from part II: Number of ways to select 4 questions from 5 = Number of ways to select 4 questions from 7 = Number of ways to select questions = = 5!/4!(5 - 4)! × 7!/4!(7 - 4)! = 5!/4!1! × 7!/4!3! = 5 × 4!/4! × 7 × 6 × 5 × 4!/4!3! = 5 × 7 × 6 × 5/(3 × 2) = 175 When a student chooses 5 questions from part I and 3 questions from part II: Number of ways to select 5 questions from 5 = Number of ways to select 3 questions from 7 = Number of ways to select questions = = 5!/5!(5 - 5)! × 7!/3!(7 - 3)! = 5!/5!0! × 7!/4!3! = 7 × 6 × 5 × 4!/4!3! = 7 × 6 × 5/(3 × 2) = 35 Hence, total number of ways in which a student can choose the questions = 210 + 175 + 35 = 420
We know that a deck of cards has 4 kings. Number of ways to select 1 king card from 4 king cards = Remaining cards in the deck = 52 - 4 = 51 Remaining number of cards to be selected = 5 - 1 = 4 Number of ways to select 4 cards from 48 cards = Total number of ways for selecting 5-card combinations = = 4!/1!(4 - 1)! × 48!/4!(48 - 4)! = 4!/3! × 48!/4!44! = 4 × 3!/3! × 48 × 47 × 46 × 45 × 44!/4!44! = 4 × 48 × 47 × 46 × 45/(4 × 3 × 2) = 778320 Hence, total number of 5-card combinations that have exactly one king = 778320
Women should occupy even places. Therefore, women will sit on places 2, 4, 6, and 8 while men will sit on places 1, 3, 5, 7, and 9. Number of ways to seat 4 women in 4 places = Number of ways to seat 5 men in 5 places = Total number of ways to seat the men and women = = 4!/(4 - 4)! × 5!/(5 - 5)! = 4!/0! × 5!/0! = 4 × 3 × 2 × 5 × 4 × 3 × 2 = 2880 Hence, total number of arrangements to seta men and women such that women that even places = 2880
Remaining students in the class = 25 - 3 = 22 Remaining number of students to be chosen for the party = 10 - 3 = 7 Number of ways to choose 7 students from 22 students = = 22!/7!(22 - 7)! = 22!/7!15! = 22 × 21 × 20 × 19 × 18 × 17 × 16 × 15!/7!15! = 22 × 21 × 20 × 19 × 18 × 17 × 16/(7 × 6 × 5 × 4 × 3 × 2) = 170544
Remaining students in the class that can go = 25 - 3 = 22 Number of students to be chosen for the excursion party = 10 Number of ways to choose 10 students from 22 students = = 22!/10!(22 - 10)! = 22!/10!12! = 22 × 21 × 20 × 19 × 18 × 17 × 16 × 15 × 14 × 13 × 12!/10!12! = 22 × 21 × 20 × 19 × 18 × 17 × 16 × 15 × 14 × 13/(10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2) = 646646 Hence, the total number of ways in which the students can be chosen for the excursion party = 170544 + 646646 = 817190
Total number of letters in the word ASSASSINATION = 13 Unique letters = {A, S, I, N, T, O} A occurs thrice, S occurs four times, I occurs twice, N occurs twice, T occurs once, and O occurs once. If the four S's are always together they can be treated as 1 letter whose position affects the permutations. Thus, remaining number of letters = 13 - 4 + 1 = 10 Number of permutations where fours S's are together = 10!/3!2!2! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3!/(3! × 2 × 2) = 151200 Hence, total number of ways to arrange the letters in the word ASSASSINATION such that the fours S's are always together = 151200 Next TopicClass 11 Maths Chapter 8 |