Class 12 Mathematics Permutation and Combination Notes and Important Questions

Class 12 Permutation and Combination Notes (Mathematics)

Algebra UNIT 1: ALGEBRA Exercise 1.1 1. A football stadium has four entrance gates and nine exists. In how many different ways can a man enter and leave the stadium? Soln: The man can enter the stadium in 4 different ways and can leave the stadium in 9 different ways. Now, bybasic principle of counting, he can enter and leave the stadium in 9 × 4 = 36 ways. 2. There are six doors in a hostel. In how many ways can a student enter the hostel and leave by a different door? Soln: The student can enter in the hostel in 6 different ways and leave in 5 different ways. Now, by the basic principle of counting, he can enter and leave by 6×5 = 30 ways. 3. In how many ways can a man send three of his children to seven different colleges of a certain town? Sonn: The man can send one of his children to 7 different colleges, next to 6 colleges and next to 5 colleges. Now, by the basic principle of counting, three of his children can send to seven different colleges of a certain town by 7 × 6 × 5 = 210 ways 4. Suppose there are five main roads between the cities A and B. In how many ways can a man go from a city to the other and return by a different road? Soln: The man has 5 option while going from a city to other and 4 while returning. Now, by the basic principle of counting, the no of ways that a man can go from a city to other and return by a different is 5 × 4 = 20 5. There are five main roads between the cities A and B and 4 between B and C. In how many ways can person drive from A to C and return without driving on the same road twice? Soln: The man has 5 option while going from city A to B and 4 option while going from B to C. Similarly, man has 3 option while returning from C to B and 4 option while returning from B to A. Now, by the basic principle of counting, total no. of ways of driving on the road = 5× 4 × 3 ×4 =240 6. How many numbers of at least three different digits can be formed from the integers 1, 2, 3, 4, 5, 6,? Soln: The numbers are of the three digits, four digits, five digits and six digits. Now, by the basic principle of counting, three-digit numbers can be formed in 6 × 5 × 4 = 120 ways, four-digit numbers can be formed in 6 × 5 × 4 × 3 = 360 ways,

*2* Solution Manual to Basic Mathematics Algebra five-digit numbers can be formed in 6× 5 × 4 × 3 × 2 = 720 ways. six-digit numbers can be formed in 6 × 5 × 4×3×2×1 = 720 ways.  Total no. of ways for selecting the numbers of at least 3 digits = 120 + 360 + 720 + 720 = 1920. 7. How many numbers of three digits less than 500 can be formed form the integers 1, 2, 3, 4, 5, 6? Soln: The required three digit numbers less than 500 should be started from 1 or 2 or 3 or 4. So, hundreds place have 4 choices. Then tens and unit place have 5 and 4 choices  Total no. of ways of forming the numbers less than 500 = 4 × 5 × 4 = 80. 8. Of the numbers formed by using all the figures 1, 2, 3, 4, 5 only once, how many are even? Soln: The last place of the even number has only two choices (2 or 4) and remaining other 4 places have respectively 4, 3, 2 and 1 choices. Now, by the basic principle of counting, no. of ways forming even numbers = 2 × 4 × 3 × 2 × 1 = 48. 9. How many numbers of different digits between 4000 and 5000 can be formed with the digits 2, 3, 4, 5, 6, 7? Soln: The number between 4000 and 5000 has 4 digits starting with 4 only. So, the thousand place has only 1 choice (i.e. 4 only) and remaining three places has respectively 5, 4 and 3 choices. Now, by the basic principle of counting, required no. of ways = 1 × 5 × 4 × 3 = 60 10. How many numbers of three different digits can be formed form the integers 2, 3, 4, 5, 6? How many of them will be divisible by 5? Soln: To form three digit numbers from the given digit. Hundreds place have 5, tens place have 4 and ones place have 3 choices. Now, by the basic principle of counting, total no. of ways to form 3-digit number = 5 × 4 × 3 = 60 Again, for the numbers divisible by 5, the last digit of the numbers should be either 0 or 5. Therefore ones place have 1 choice and remaining two places has 4 and 3 choices. Now, by the basic principle of counting, required no. of ways = 1× 4 × 3 = 12 ways. Hint and Solution of MCQ’s 1. The man has 6, 5 and 4 different college choices for his first, second and third son  the man can send 3 of his child by 6  5  4 = 120 ways 2. The man has 8 choices from Kathmandu to Pokhara and he has 7 choices to return back.  the man has 8  7 = 56 way to go from Kathmandu to Pokhara and return back. 3. To form 2-digit numbers with the integers 3, 4, 5, 6 and 7 Ones place has 5 different choices Tens place has 4 different choices  the two different digit numbers can be formed by 5  4 = 20 ways.

Permutation and Combination*3* Algebra 4. To form 2-digit numbers with the integers 1, 2, 3, 4, 5 and 6 Ones place has 6 different choices Tens place has 6 different choices  the two different digit numbers can be formed by 6  6 = 36 ways. 5. To form the numbers between 100 and 1000 such that every digit is either 2 or 9, we have to form 3- digit numbers using the integers. Ones place has 2 choices Tens place has 2 choices Hundreds place has 2 choices  there are 2  2  2 = 8 ways. 6. To wear 1 pant, 1 shirt and 1 tie, he has the choice of 6 pants, 5 shirts and 4 ties.  there are 6  5  4 = 120 different outfit to go the college. 7. To form 3-different digit odd numbers with the integers 5, 6, 7, 8, 9 Ones place has 3 different choices (either 5, 7 or 9) Tens place has 4 different choices Hundreds place has 3 different choices  the 3-different digit odd numbers can be formed by 3  4  3 = 36 ways. 8. To answer each 6 true or false questions, the man has 2 choices in each  the number of possible answers altogether is 26 = 64. Exercise 1.2. 1. Find the number of permutations of five different object taken three at a time. Soln: The permutation of n = 5 objects taken r = 3 objects can be made by P(n, r) = P(5, 3) = 5! (5–3)! = 5! 2! = 5 × 4× 3 = 60 ways. 2. If three persons enter a bus in which there are ten vacant seats, find in how many ways they can sit. Soln: Three passenger can be arrange in 10 seat by P(n, r) = P (10, 3)= 10! (10–3)! = 10! 7! = 10×9×8×7! 7! = 720 ways 3. a) How may plates of vehicles consisting of 4 different digits can be made out of the integers 4, 5, 6, 7, 8, 9?How many of these numbers are divisible by 2? Soln: From the 6 integers 4 different digit numbers can be formed by P(n, r) = P(6, 4) = 6! (6–4)! = 6! 2! = 6×5×4×3×2×1 2×1 = 360 For divisible by 2; Since the even numbers of 4 digit numbers are divisible by 2. So, the unit place of the number can arranged in 3 ways.Then, the remaining 3 digits of the number can be arranged from 5 integers in P(5, 3) ways.Now, by the basic principle of counting,  The required number of 4 digits divisible by 2 = 3 × P(5, 3) = 3 × 5! (5 – 3)! = 180

*4* Solution Manual to Basic Mathematics Algebra b) How many numbers of 4 different digits can be formed from the digits 2, 3, 4, 5, 6, 7? How many of these numbers are; i) divisible by 5 ii) not divisible by 5. Soln: Here, total no. of digits 2, 3, 4, 5, 6, 7, n = 6, no. of taken digits for a number, r = 4 Now, by the definition of permutation, The required number of permutations, P(n, r) = P(6, 4) = 6! (6–4)! = 6×5×4×3×2×1 2×1 = 360 i) For divisible by 5, once place is 5 in the given digits that can be arranged in 1 way. Other 3 digits can be arranged in P(5, 3) ways.Now, by the basic principle of counting, required number of 4 digits divisible by 5 = 1 × P(5, 3) = 5! (5 – 3)! = 5×4×3×2! 2! = 60 ii) The number can be arranged not divisible y 5 = 360 – 60 = 300 c) How many 5 digit odd numbers can be formed using the digits 3, 4, 5, 6, 7, 8 and 9 if; i) repetition of digits is not allowed ii) repetition of digits is allowed? Soln: i) To form 5 digit odd number form the digits 3, 4, 5, 6, 7, 8 and 9when repetition not allowed, The arrangement ofunit place has 4 choices (3 or 5 or 7 or 9) and remaining 4 places can be arranged in P(6, 4) = 6! (6–4)! = 6! 2! = 360 ways. The required number of arrangement if repetition is not allowed = 4  360 = 1440 ways. ii) To form 5 digit odd number form the digits 3, 4, 5, 6, 7, 8 and 9 and repetition allowed, The arrangement ofunit place has 4 choices (3 or 5 or 7 or 9) and remaining each 4 places can be arranged in 7 ways.  The number of arrangements if repetition is allowed = 4  7  7  7  7 = 9604 ways. 4. In how many ways can four boys and three girls be seated in a row containing seven seats? a) if they may sit anywhere b) if the boys and girls must alternate c) if all three girls are together? d) if girls are to occupy odd sets. Soln: There are 4 boys and 3 girls. Then total no. of persons (n) = 4 + 3 = 7. Total no. of seats in row (r) = 7 a) The number of arrangements of 7 person in a row = P (7, 7) = 7! (7 – 7)! = 7! 0! = 5040 ways. b) To arrange the boys and girls alternately 4 boys can be arrange in 4 seatsby P (4, 4) = 4! = 4×3× 2 × 1 = 24 ways 3 girls can be arrange in 4 seatsbyP (3, 3) = 3! = 3×2 × 1 = 6 ways.  Required ways = 24 × 6 = 144 c) As all three girls are together, let's consider them as single unit, the number of ways that 3 girls together and 4 boys be seated in a row = P (5, 5) = 5! = 5 × 4 × 3 × 2 ×1 = 120 ways Also, three girls sitting together can sit in p (3, 3) = 3! = 3 × 2 ×1 = 6 ways  Required ways = 120  6 = 720

Permutation and Combination*5* Algebra d) If girls are to occupy odd sets, the 3 girls can be arrange 4 odd seats by P(4, 3) = 4! = 24 ways Again, remaining 4 boys can be arrange in remaining 4 seats by P(4, 4) = 4! = 24 ways  Required arrangements = 24  24 = 576 ways. 5. In how many ways can eight people be seated in a row of eight seats so that two particular persons are; a) always together b) never together? Soln: Here, the arrangements of 8 people in a row is 8! = 40320 ways a) When two particular persons are always together, we have to consider these two people as one. Now, no. arrangements of 7 people on 7 seats = P(7, 7) = 7! = 5040 ways The two persons kept together has their P (2, 2)= 2! arrangements  The required arrangements = P (2, 2) × P (7, 7) = 2! × 7! = 10080 ways. b) When two particular persons are never together,  The required arrangements =40320 – 10080 = 30240 ways. 6. Six different books are arranged on a shelf. Find the number of different ways in which two particular books are; a) always together b) not together. Soln: a) Let's suppose two books as single unit. Then, no of ways in which two particular books are always together = p (5, 5) = 5! The two books kept together has their P (2, 2)= 2! arrangements  Number of required arrangement 5! × 2! = 240 ways. b) Total no. of arrangement of books = P (6, 6) = 6! = 720 ways. No. of arrangement in which two particular books are not together = 720 – 240 = 480 ways. 7. In how many ways can four red beads, five white beads and three blue beads be arranged in a row? Soln: Total no. of beads = 12, in which red beads = 4, white beads = 5 and blue beads = 3.  Required no. of permutations = 12! 4! 5! 3! = 8. In how many ways can the letters of the following words be arranged? a) ELEMENT b) NOTATION c) MATHEMATICS d) MISSISSIPPI Soln: a) Total no. of letters in the word “ELEMENT” is 7 no. of repetition of E = 3  Required no of arrangements = 7! 3! =7 × 6 × 5 × 4 × 3! 3! = 840 b) Total no. of letters in the word “NOTATION” is 8 no. of repetition of N= 2 no. of repetition of O = 2 no. of repetition of T= 2  Required no. of arrangements = 8! 2! 2! 2! = 8  7 × 6 × 5 × 4 × 3  2! 2  2  2! = 5040

*6* Solution Manual to Basic Mathematics Algebra c) Total no. of letters in the word "MATHEMATICS" is 11 no. of repetition of M= 2 no. of repetition of A = 2 no. of repetition of T= 2  Required no. of arrangements = 11! 2! 2! 2! d) There are 11 letters in the given word "MISSISSIPPI" is 11 no. of repetition of I = 4 no. of repetition of S = 4 no. of repetition of P = 2  Required no. of arrangements = 11! 4!4!2! 9. How many numbers of 6 digits can be formed with the digits 2, 3, 2, 0, 3, 3? Soln: The given digits are 2, 3, 2, 0, 3, 3. Total no. of digits (n) = 6 no. of repetition of 2 = 2 no. of repetition of 3 = 3 Then the number ways of forming 6 digits number = 6! 2!3! = 60 The number of 6 digits numbers with 0 in the beginning = 1 5! 2!3!  = 5.4.3.2.1 2.1 × 3.2.1 = 10  The required no. of 6 digits number = 60 – 10 = 50 10. In how many ways can 4 art students and 4 Science students be arranged in circular table if; a) they may sit anywhere b) they sit alternately. Soln: a) Total number of students (n) = 4 + 4 = 8 If they sit anywhere, the no. of circular arrangements of 8 students = (8 – 1)! = 7! = 5040 ways b) If science and art students sits alternately in a round table, The 4 art students may sit in their 4 position of a round table by (4 – 1)! = 3×2×1 = 6 ways and The 4 science students may sit in their 4 seats by 4! = 4×3×2×1= 24 ways.  The required no. of arrangements = 6 × 24 = 144 ways 11. In how many ways can eight people be seated in a round table if two people insist in sitting next to each other? Soln: Number of people (n) = 8 If two people have to insist on sitting next to each other, these two people are considered as one. Then we have to arrange 7 people in round table. Which can be done by (7 – 1)! = 6! = 720 ways Again, these two people can interchange their position in 2! = 2 ways.  The required no. of arrangements = 2 × 720 = 1440 ways.

Permutation and Combination*7* Algebra 12. In how many ways can seven different colored beads be made into a bracelet? Soln: Number of beads (n) = 7 The no. of arrangements of 7 beads in a circle = (7 – 1)! = 6! = 720 ways In case of bracelet, there is not significant difference in clockwise and anticlockwise arrangement.  The required no. of arrangements = 1 2 × 720 = 360 ways. 13. a) In how many ways can 4 letters be posted in six letter boxes? Soln: No. of boxes, (n) = 6 No. of letters (r) = 4  4 letters can be posted in 6 letter box by nr = 64 = 1296 ways. b) How many even numbers of 3 digits can be formed when repetition of digits is allowed? Soln: The even numbers of three digits is to form using the digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 Here, repetition of digits is allowed to form 3 digit numbers, so Unit place of the number has 5 choices (0, 2, 4, 6, or 8) Tens place of the numbers has 10 choices (0, 1, 2, 3, 4, 5, 6, 7, 8 or 9) Hundreds place of the number has 9 choices (1, 2, 3, 4, 5, 6, 7, 8, 9)  Required no. arrangements = 9 × 10 × 5 = 450 ways. c) In how many ways can 3 prizes be distributed among 4 students so that each student may receive any number of prizes? Soln: Here, no. of students (n) = 4, no. of prizes (r) = 3 The required no. of ways that 3 prizes be distributed among 4 students = 43 = 64 ways. 14. In how many ways can the letters of the word "MONDAY" be arranged? How many of these arrangements do not begin with M? How many begin with M and do not end with Y? Soln: Here,Total no. of letters in the word "MONDAY" is 6 The number of arrangements of 6 letters = 6! = 720 ways. The number of arrangement, which begins from M is = 1 × 5! = 120 ways.  Required no. of arrangements which do not begin with M = 720 – 120 = 600 ways. The number of arrangements, which begins with M and end with Y = 1 × 4! × 1= 24 ways Required no of arrangements, which begins from M and do not end with Y = 120 – 24 = 96 ways. 15. Show that the number of ways in which the letters of the word a) "COLLEGE" can be arranged so that the two E's always come together is 360. Soln: The number of letters in the word COLLEGE is 7 There are 2L’s and 2 E's. When 2 E's come together, these 2E's is considered as one unit. Then, number of letters = 6 and contains 2L’s  The required number of arrangements = 6! 2! = 360 ways. b) "ARRANGE" can be arranged so that no 2 R's come together is 900. Soln: The number of letters in the word 'ARRANGE' is 7 The number of permutation = 7! 2!2! = 7×6×5×4×3×2×1 2×1 × 2×1 = 1260 ways

*8* Solution Manual to Basic Mathematics Algebra There are 2A's and 2 R's.When 2 R's come together, these 2R's is considered as one unit. Then, number of letters = 6 and contain 2A's So, number of permutation in which 2R's come together = 6! 2! = 6.5.4.3.2.1 2.1 = 360 ways  The required number of arrangements 2R's do not come together= 1260 – 360 = 900 ways. 16. In how many ways can the letters of the word "COMPUTER" be arranged so that (a) all the vowels are always together? Soln:The word “COMPUTER", has three vowels (O, U,E) and 5 consonants (C, M, P, T, R). When vowels are together,then we have 6 letters.  The no. of ways of their arrangement = 6! Also, the vowels can change their position in 3! ways  Total no. of arrangements, so that all vowels are always together = (6!) × (3!) = 4320 ways. b) the vowels may occupy only odd positions? Soln: Here, three vowels can be arrange in 4 odd position by P(4, 3) ways and remaining consonants can be arranged in 5! ways.  The required number of arrangements = P(4, 3)×5! = 4! (4 –3)!×5!= 2880 ways. c) the relative positions of vowels and consonants are not changed? Soln: The 3 vowels are arranged in 3 places by 3! = 6 ways and The remaining 5 consonants are arranged in 5 places by 5! = 120 ways.  Required no. of ways = 6 × 120 = 720 ways. 17. Find the number of arrangements of the letters of the word "LAPTOP" so that; a) the vowels may never be separated, b) all consonants may not be together, c) they always begin with L and end with T, d) they do not begin with L but always end with T. Soln: The word “LAPTOP” has two vowels(A, O) and four consonants (L, P, T, P) a) When vowels may never be separated, then no. of letters to arrange is 5 containing 2P’s.  No. of arrangementsvowels together =5! 2! = 5  4  3  2! 2! = 60 ways. Again, two vowels can be arranged in P(2, 2) = 2! = 2 ways.  Required no. of arrangements = 60  2 = 120 ways. b) The number of arrangements of letters of the word = 6! 2! = 360 ways. When consonants are altogether, then we have to arrange 3 letters considering all consonants as one  No. of arrangements of 3 letters = 3! = 6 ways. Also, fourconsonants can be arranged in P(4, 2) = 12 ways.  The no. of arrangements of consonant together = 6  12 = 72 ways. Hence, required no. of arrangements if all consonants may not be together = 360 – 72 = 288 ways.

Permutation and Combination*9* Algebra c) When they always begin with L and end with T, The no. of arrangements = 1 4! 2!  1 =12 ways. d) When they do not begin with L but always end with T, The beginning position has 4 choices (A, P, T, O), last position has 1 choice (T), and remaining 4 position can be arranged in 4! 2! ways, as there are 2P’s  The no. of arrangements that do not begin with L and end with T = 4 4! 2! 1 = 48 ways. 18. How many different words can be formed with all the letters of the word "Internet" if; a) each word is to begin with vowel? b) each word is to end with consonant? Soln: Here, the word “INTERNET” has 8 letters in which 3 vowels (I, E, E) and 6 consonants (N, N, R, T, T) a) When the arrangements begin with I, n = 7, p = 2, q = 2, r = 2.  Total no. of arrangementsbegin with I= n! p! q! r! = 7! 2!2!2! = 7  6  5  4  3  2! 2  1  2  1 2! = 630 ways. When the arrangements begin with E, n = 7, p = 2, q = 2.  No. of arrangements begin with E = n! p! q! = 7! 2!2! = 7  6  5  4  3  2! 2  1 2! = 1260 ways. Hence, total of arrangements begin with vowels = 630 + 1260 = 1890 ways. b) When the arrangements end with consonants = total no. of arrangements – arrangements end with vowel = total no. of arrangements – arrangements begin with vowel = 8! 2!2!2! –1890 = 5040 – 1890 = 3150 Hint and Solution of MCQ’s 1. P(n, n – r) =   !r ! n ! r n n ! n    2. Since P(n, n) = n!, P(6, 6) = 6! 3. 25 ! 4 5 ! 4 5 ! 4 ) 1 6 ( ! 5 ! 4 ! 5 ! 5 6 ! 4 ! 5 ! 6           4. The number of n identical objects taken r at a time is 1. 5. P(n, 4) = 2P(n, 3)   ! 3 n ! n

)! 4 n ( ! n     3 n

2 1    n – 3 = 2  n = 5 6. n! = 56(n – 2)!  n(n – 1) = 56  n(n – 1) = 8 ( 8 – 1)  n = 8

*10* Solution Manual to Basic Mathematics Algebra 7. No. of integers = 5 The 3 different digits license plate from the five different integers can be formed by

)! 3 5 ( ! 5 ) 3 , 5 ( P    8. No. of digits = 4 To form 3 digit numbers, each three positions has four choices  the number of ways in which here different digit with repetition is 43. 9. The number of hoisting 2 blue, 2 white and 2 yellow flags on the pole at the same time is ! 2 ! 2 ! 2 ! 6 . 10. The letters of the word SMALL can be arranged in 60 ! 2 ! 5  ways. 11. The letters to arrange to have L together ( H, O, LL,O,W) is 5 in which O’s are 2  the number of arrangements = 60 ! 2 ! 5  12. The number of ways 6 different beads can be strung on the necklace = 60 )! 1 6 (

)! 1 n (

1     13. No. of gentleman = 4 and No. of ladies = 3 To find the arrangements no ladies are together, they must be alternately being any two boys always together. So, 4 gentlemen can be arrange in their position independently by 4! and 3 ladies can be arrange in their position independently by 3! ways.  the number of arrangements no ladies are together = 4!  3! ways. Exercise 1.3. 1. A boy puts his hand into a bag which contains 10 differently coloured marbles and brings out 3. How many different results are possible? Soln:The 3 marbles out of 10 differently coloured marbles can be bring in C(10, 3) = 120 different ways. 2. Find the number of ways in which a student can select 5 courses out of 8 courses. If 3 courses are compulsory, in how many ways can the selections be made? Soln: The student can select 5 courses out of 8 courses in C (8, 5) = 8! 5! (8 – 5)! = 56ways. If 3 courses are compulsory, students have to select (5 – 3) = 2 courses out of (8 – 3) = 5.  The student can select 5 course out of 8 courses if three courses are compulsory by C (5, 2) = 5! (5–2)! 2! = 10ways 3. From 10 persons, in how many ways can a selection of 4 be made i) when one particular person is always included? ii) when two particular persons are always excluded? Soln: i) When one particular person is always included, we have to choose 3 persons out of 9  No. of ways of selection = C (9,3) = 9! (9–3)! (3!) = 84

Permutation and Combination*11* Algebra ii) When two particular persons are always excluded we have to choose 4 persons out of 8  No. or ways of selection = C(8, 4)= 8! (8 – 4)! (4!) =70 4. A bag contains 8 white balls and 5 blue balls. In how many ways can 5 white balls and 3 blue balls be drawn? Soln: The 5 white balls out of 8 white can be select in C (8, 5) ways and The 3 blue balls out of 5 can be select in = C(5, 3) ways  Required no. of selection = C(8, 5) × C(5, 3) = 56× 10 = 560 ways. 5. How many committees can be formed from a set of 7 boys and 5 girls if each committee contains 4 boys and 3 girls? Soln: The 4 boys our of 7 can be select in C(7, 4) ways and The 3 girls out of 5 can be select in C(5, 3) ways  Required no. of committees = C (7, 4) × C(5, 3)= 35 × 10 = 350. 6. From a group of 11 men and 8 women, how many committees consisting of 3 men and 2 women are possible? Soln: The 3 boys our of 11 can be select in C(11, 3) ways and The 2women out of 8 can be select in C(8, 2) ways  Required no. of committees = C (11, 3) × C(8, 2)= 165 × 28 = 4620. 7. From 4 mathematician, 6 statistician and 5 economists, how many committees of 6 members can be formed so as to include 2 members from each category? Soln: To make a committee of 6 members inducing 2 members from each category we have to made a selection of 2 mathematician out of 4, 2 statistician out of 6 and 2 economists out of 5  Required no. of committees = C(4,2) × C(6,2) × C(5,2)= 6 × 15 × 10 = 900. 8. A person has got 12 acquaintances of whom 8 are relatives. In how many ways can he invite 7 guests so that 5 of them may be relatives? Soln: The person has to invite 7 guests of whom 5 are relatives from 8 relatives and rest of 2 non-relatives from 5 non-relatives.  Required no. of invitations = 8C5×4C2= 56 × 6 = 336ways. 9. There are ten electric bulbs in the stock of a shop out of which there are three defectives. In how many ways can a selection of 6 bulbs be made so that 4 of them may be good bulbs? Soln: The selection of 6 bulbs in which 4 good bulbs is to select from 7 good bulbs and remaining 2 defective from 3 defective bulbs  Required no of ways = C(7, 4) × C(3, 2) = 35×3 = 105. 10. From 6 gentlemen and 4 ladies, a committee of 5 is to be formed. In how many ways can this be done so as to include at least one lady? Soln: Theways of selection of 5 people from 6 gentlemen and 4 ladies including at least 1 lady is given by Ladies (4) Gentlemen (6) Selection 1 4 C (4, 1) × C (6, 4) 2 3 C (4, 2) × C (6, 3) 3 2 C (4, 3) × C (6, 2) 4 1 C (4, 4) × C (6, 1)

*12* Solution Manual to Basic Mathematics Algebra  Total ways of selection = C (4, 1) × C (6, 4) + C (4, 2) × C (6, 3) + C (4, 3) × C (6, 2) +C (4, 4) × C (6, 1) = 4 × 15 + 6 × 20 + 4 × 15 + 6 = 60 + 120 + 60 + 6 = 246 ways. 11. A candidate is required to answer 6 out of 10 questions which are divided into 2 groups each containing 5 questions and he is not permitted to attempt more than 4 from any group. In how many different ways can he make up his choice? Soln: The ways of selection 6 questions including not more than 4 from each group of containing 5 questions is given by Group A(5) Group B(5) Selection 4 2 C (5, 4) × C (5, 2) 3 3 C (5, 3) × C (5, 3) 2 4 C (5, 2) × C (5, 4)  Total number of selection = C (5, 4) × C (5, 2) + C (5, 3) × C (5, 3) + C (5, 2) × C (5, 4) = 50 + 100 + 50= 200 ways. 12. A man has 5 friends. In how many ways can he invite one or more of them to a dinner? Soln: The mancan invite 1 out of 5 or 2 out of 5 or 3 out of 5 or 4 out of 5 or 5 out of 5 to a dinner  No. of ways invitation = C(5,1) + C(5,2) + C(5,3) + C(5,4) + C(5,5) = 5 + 10 + 10 + 5 + 2= 31. 13. a) If C(20, r + 5) = C(20, 2r – 7), find C(15, r). Soln: SinceC(n, r) = C(n, r')  r = r' or r + r' = n,  C(20, r + 5) = C(20, 2r - 7)  r + 5 = 2r – 7 or r + 5 + 2r – 7 = 20  r = 12 or r = 22 3 (not possible)  r = 12  C(15, r) = C(15, 12) = 15! 12! (15 - 12)! = 455. b) If C(n, 10) + C(n, 9) = C(20, 10) find n and C(n, 17) Soln: SinceC(n, r) + C(n, r + 1) = (n + 1, r + 1), C (n, 10) + C (n, 9) = C(20, 10)  C(n + 1, 10) = C(20, 10)  n + 1= 20  n = 19  C (n, 17) = C(19, 17) = 19! 2! 17! = 171. c) Solve for n the equation C(n+ 2, 4) = 6C(n, 2) Soln: C(n+2,4) = 6 C (n, 2).  (n + 2)! (n + 2 –4)! 4! = n! (n–2)! 2!  (n + 2) (n + 1) n! 4.3.2! = 6. n! 2!

Permutation and Combination*13* Algebra  (n + 2) (n + 1) = 72  n2 + 3n – 70 = 0,  (n + 10)(n – 7) = 0  n = 7 or n = - 10  n = 7 (neglecting the negative number). d) If P(n, r) = 336 and C(n, r) = 56, find n and r. Soln: P(n, r) = r! C(n, r)  336 = r!  56  r! = 6 = 3!  r = 3. P(n, r) = n! (n–r)!  336 = n(n – 1)(n – 2)(n – 3)! (n–3)!  8  7  6 = n(n – 1)(n – 2)  8  (8 – 1)  (8 – 2) = n(n – 1)(n – 2)  n = 8. e) If nCr– 1 = 45, nCr = 120 and nCr+ 1 = 210, find n and r. Soln: Here,nCr– 1 = 45, nCr = 120 and nCr+ 1 = 210 Now, nCr– 1 = 45  n! (n–r + 1)!(r – 1)! = 45 ……. (i) nCr = 120  n! (n–r)! r! = 120 ……………….. (ii) nCr+ 1 = 210  n! (n–r – 1)! (r + 1)! = 210 ……. (iii) Dividing eqn (i) by (ii), we get

120 = n! (n–r + 1)!(r – 1)! (n–r)! r! n!  3 8 = (n–r)! r(r – 1)! (n – r + 1)(n – r)! (r – 1)!  3 8 = r n – r + 1  3n – 3r + 3 = 8r  3n + 3 = 11r ………. (iv) Again, dividing eqn (iii) by (ii), we get

120 = n! (n–r – 1)!(r + 1)! (n–r)! r! n!  7 4 = (n–r)(n – r – 1)! r! (n – r – 1)! (r + 1) r!  7 4 = n–r r + 1  4n – 4r = 7r + 7  4n – 7 = 11r ………. (v)

*14* Solution Manual to Basic Mathematics Algebra From eqn (iv) and (v), we get 3n + 3 = 4n – 7  n = 10. Substituting the value of n in eqn (iv), we get 4  10 – 7 = 11r  r = 3. 14. An examination paper consisting of 10 questions, is divided into two groups A and B. Group A contains 6 questions. In how many ways can an examinee attempt 7 questions; a) selecting 4 from group A and 3 from group B? b) selecting at least two questions from each group? Soln: Total no. of questions = 10, No. of questions in group A = 6  No. of questions in group B = 10 – 6 = 4 No. of questions to be attempt = 7 a) The number of selection of 4 questions from 6 questions = C(6, 4) The number of selection of 3 questions from 4 questions = C(4, 3)  Required no. of selection = C(6, 4) × C(4, 3) = 15  4 = 60 ways. b) The ways of selection of 7 questions: Group A (6) Group B (4) Selection 5 2 C(6, 5) × C(4, 2) 4 3 C(6, 4) × C(4, 3) 3 4 C(6, 3) × C(4, 4)  Total number of selection = C(6, 5) × C(4, 2) + C(6, 4) × C(4, 3) + C(6, 3) × C(4, 2) = 36 + 60 + 20= 116 ways. 15. Six men in a group of 8 are skilled. Find the number of ways by which 5 men can be selected such that a) at least 3 of them may be the skilled men. b) at least one of them may be the unskilled man. Soln: No. of men= 8, No. of skilled men = 6, No. of unskilled men = 8 – 6 = 2, No. of men to be selected (r) = 5 a) Theways of selection of 5 men consisting at least 3 of them may be the skilled: Skilled men (6) Unskilled men(2) Selection 3 2 C(6, 3) × C(2, 2) 4 1 C(6, 4) × C(2, 1) 5 0 C(6, 5) × C(2, 0)  Total number of selection = C(6, 3) × C(2, 2) + C(6, 4) × C(2, 1) + C(6, 5) × C(2, 0) = 20 + 30 + 6= 56 ways. b) Theways of selection of 5 men consisting at least1 of them may be the unskilled: Skilled men (6) Unskilled men(2) Selection 4 1 C(6, 4) × C(2, 1) 3 2 C(6, 3) × C(2, 2)

Permutation and Combination*15* Algebra  Total number of selection = C(6, 4) × C(2, 1) + C(6, 3) × C(2, 2) = 30 + 20 = 50 ways. 16. In a group of 10 students, 6 are boys. In how many ways can 4 students be selected for mathematical competition so as to include; a) exactly two boys b) at least two boys c) at most two girls Soln: Total no. of students = 10, No. of boys = 6, No. of girls = 10 – 6 = 4, No. of students to select = 4 a) The selection of selecting exactly 2 boys from 6 boys and 2 girls from 4 girls = C(6, 2) × C(4, 2)= 15 × 6= 90 ways b) The ways of selection of 4 students consisting at least 2 boys: Boys (6) Girls (4) Selecting ways 2 2 C(6, 2) × C(4, 2) 3 1 C(6, 3) × C(4, 1) 4 0 C(6, 4) × C(4, 0)  Total number of selection = C(6, 2) × C(4, 2) + C(6, 3) × C(4, 1) + C(6, 4) × C(4, 0) = 90 + 80 + 15 = 185 ways. c) The ways of selection of 4 students consisting at most 2 girls, Girls (4) Boys (6) Selecting ways 2 2 C(4, 2) × C(6, 2) 1 3 C(4, 1) × C(6, 3) 0 4 C(4, 0) × C(6, 4) Total number of selection = C(4, 2) × C(6, 2) + C(4, 1) × C(6, 3) + C(4, 0) × C(6, 4) = 90 + 80 + 15 = 185 ways. Hint and Solution of MCQ’s 1. All the relations given in a), b), c) are true 2. C(n, n – 1) = C(n, 1) = n [C(n, n – r) = C(n, r)] 3. Since P(n, r) = r! C(n, r), P(n, r) = C(n, r)  r ! = 1  r = 0 or 1 4. C(n , 5) + C(n, 4) = C(n + 1, 5) [C(n, r) + C(n, n – 1) = C(n + 1) , r] 5. C(20, r + 1) = C(20, 3r – 1)  r + 1 = 3r – 1 or r + 1 + 3r – 1 = 20  r = 1 or r = 5 6. P(4, 2) = n C(4, 2)  2! C(4, 2) = n C(4, 2) [P(n, r) = r! C(n, r)]  n = 2! = 2 7. P(n, r) = r! C(n, r)  60 = r! 10  r! = 6  r! = 3!  r = 3

*16* Solution Manual to Basic Mathematics Algebra 8. If 1 question is compulsory, the student has to chose 5 questions out of 7. Which can be done by C(7, 5) = 21 ways 9. No. of teams = 12 If each team has to play with other, the no. of matches played = C(12, 2) = 66 10. Total person = 5 gentlemen + 4 ladies = 9 Out of 9 person, a committee of 6 members can be formed by C(9, 6) = 84 ways 11. When m players are excluded, the selection of r player has to made from n – m player. Which can be done by C(n – m, r) ways. 12. When two particular persons are always included, the selection of 5 – 2 = 3 person has to made from 8 – 2 = 6 person. Which can be done by C(6, 3) = 20 ways. 13. From 4 coins, the different sums can be form choosing one, two, three or four coins. Which can be done by C(4, 1) + C(4, 2) + C(4, 3) + C(4, 4) = 4 + 6 + 4 + 1 = 15 ways. 14. The man can invite one, two, three , four or five of them to a dinner. Which can be done by C(5, 1) + C(5, 2) + C(5, 3) + C(5, 4) + C(5, 5) = 5 + 10 + 10 + 5 + 1 = 31 