CBSE Class 11 Mathematics Permutations and Combinations Assignment Set 02

Question. A college offers 7 courses in the morning and 5 in the evening. The number of ways a student can select exactly one course, either in the morning or in the evening:
a. 27
b. 15
c. 12
d. 35
Answer : C

Question. The number of arrangement of the letters of the word “CALCUTTA”?
a. 2520
b. 5040
c. 10080
d. 40320
Answer : B

Question. How many words can be made from the letters of the word ‘COMMITTEE?’
a. 9!/(2!)²
b. 9!/(2!)³
c. 9!/2!
d. 9 !
Answer : B

Question. If eleven members of a committee sit at a round table so that the President and Secretary always sit together, then the number of arrangements is:
a. 10! × 2
b. 10!
c. 9! × 2
d. None of these
Answer : C

Question. If the letters of the word ‘KRISNA’ are arranged in all possible ways and these words are written out as in a dictionary, then the rank of the word ‘KRISNA’ is:
a. 324
b. 341
c. 359
d. None of these
Answer : A

Question. We are to form different words with the letters of the word ‘INTEGER’. Let 1 m be the number of words in which I and N are never together, and 2 m be the number of words which begin with I and end with R. Then 1 2 m /m is equal to:
a. 30
b. 60
c. 90
d. 180
Answer : A

Question. 20 persons are invited for a party. In how many different ways can they and the host be seated at a circular table, if the two particular persons are to be seated on either side of the host?
a. 20!
b. 2.18!
c. 18!
d. None of these
Answer : B

Question. The number of numbers that can be formed with the help of the digits 1, 2, 3, 4, 3, 2, 1 so that odd digits always occupy odd places, is:
a. 24
b. 18
c. 12
d. 30
Answer : B

Question. How many numbers can be made with the help of the digits 0, 1, 2, 3, 4, 5 which are greater than 3000: (repetition is not allowed)
a. 180
b. 360
c. 1380
d. 1500
Answer : C

Question. m men and n women are to be seated in a row, so that no two women sit together. If m > n , then the number of ways in which they can be seated is:
a. m!(m + 1)! / (m -  n + 1)!
b. m!(m - 1)! / (m - n + 1)!
c. (m - 1)!(m + 1)! / (m - n + 1)!
d. None of these
Answer : A

Question. In a monthly test, the teacher decides that there will be three questions, one from each of exercise 7, 8 and 9 of the text book. If there are 12 questions in exercise 7, 18 in exercise 8 and 9 in exercise 9, in how many ways can three questions be selected?
a. 1944
b. 1499
c. 4991
d. None of these
Answer : A

Question. An n digit number is a positive number with exactly n digits. Nine hundred distinct n-digit numbers are to be formed using only the three digits 2, 5 and 7. The smallest value of n for which this is possible is:
a. 6
b. 7
c. 8
d. 9
Answer : B

Question. The number of ways in which 5 male and 2 female members of a committee can be seated around a round table so that the two female are not seated together is:
a. 480
b. 600
c. 720
d. 840
Answer : A

Question. The number of ways that 8 beads of different colours be string as a necklace is:
a. 2520
b. 2880
c. 5040
d. 4320
Answer : A

Question. In how many ways 7 men and 7 women can be seated around a round table such that no two women can sit together:
a. (7!)2
b. 7!×6!
c. (6!)2
d. 7!
Answer : B

Question. In how many ways a team of 10 players out of 22 players can be made if 6 particular players are always to be included and 4 particular players are always excluded:
a. ²²C₁₀
b. ¹⁸C₃
c. ¹²C₄
d. ¹⁸C₄
Answer : C

Question. If ¹⁵C_3r = ¹⁵C_r+3, then the value of r is:
a. 3
b. 4
c. 5
d. 8
Answer : 

Question. ⁿC_r / ⁿC_r-1 = ?
a. n - r / r
b. n + r - 1 / r
c. n - r - 1 / r
d. n - r - 1 / r
Answer : C

Question. In how many ways can a girl and a boy be selected from a group of 15 boys and 8 girls?
a. 15 x 8
b. 15 + 8
c. ²³P₂
d. ²³C₂
Answer : A

Question. There are 10 lamps in a hall. Each one of them can be switched on independently. The number of ways in which the hall can be illuminated is?
a. 10²
b. 1023
c. 2¹⁰
d. 10!
Answer : B

Question. In how many ways can 5 boys and 5 girls sit in a circle so that no two boys sit together?
a. 5!×5!
b. 4!× 5!
c. 5 ! x 5 ! / 2
d. None of these
Answer : B

Question. In how many ways can 15 members of a council sit along a circular table, when the Secretary is to sit on one side of the Chairman and the Deputy Secretary on the other side?
a. 2×12!
b. 24
c. 2×15!
d. None of these
Answer : A

Question. If ⁿ⁺¹C₃ = 2ⁿC₂, then n = ?
a. 3
b. 4
c. 5
d. 6
Answer : C

Question. The number of ways in which 5 beads of different colours form a necklace is:
a. 12
b. 24
c. 120
d. 60
Answer : A

Question. The number of ways in which 9 persons can be divided into three equal groups is:
a. 1680
b. 840
c. 560
d. 280
Answer : D

Question. A man has 7 friends. In how many ways he can invite one or more of them for a tea party:
a. 128
b. 256
c. 127
d. 130
Answer : C

Question. In a city no two persons have identical set of teeth and there is no person without a tooth. Also no person has more than 32 teeth. If we disregard the shape and size of tooth and consider only the positioning of the teeth, then the maximum population of the city is:
a. 2³²
b. (32)² −1 
c. 2³² − 1
d. 2³²⁻¹
Answer : C

Question. If : ²ⁿC₃ , ⁿC₂ = 44 : 3, then for which of the following values of r , the value of ⁿC_r will be 15?
a. r = 3
b. r = 4
c. r = 6
d. r = 5
Answer : B

Question. If ^n2-nC₂ = ^n2-nC₁₀ then n = ?
a. 12
b. 4 only
c. −3 only
d. 4 or −3
Answer : D

Question. Numbers greater than 1000 but not greater than 4000 which can be formed with the digits 0, 1, 2, 3, 4 (repetition of digits is allowed), are:
a. 350
b. 375
c. 450
d. 576
Answer : B

Question. In the 13 cricket players 4 are bowlers, then how many ways can form a cricket team of 11 players in which at least 2 bowlers included:
a. 55
b. 72
c. 78
d. None of these
Answer : C

Question. If ⁸C_r = ⁸C_r+2 , then the value of C₂ is:
a. 8
b. 3
c. 5
d. 2
Answer : B

Question. In how many ways can 5 prizes be distributed among four students when every student can take one or more prizes?
a. 1024
b. 625
c. 120
d. 60
Answer : A

Question. In an examination there are three multiple choice questions and each question has 4 choices. Number of ways in which a student can fail to get all answers correct, is:
a. 11
b. 12
c. 27
d. 63
Answer : D

Question. In an election there are 8 candidates, out of which 5 are to be choosen. If a voter may vote for any number of candidates but not greater than the number to be choosen, then in how many ways can a voter vote:
a. 216
b. 114
c. 218
d. None of these
Answer : C

Question. In a touring cricket team there are 16 players in all including 5 bowlers and 2 wicket-keepers. How many teams of 11 players from these, can be chosen, so as to include three bowlers and one wicket-keeper:
a. 650
b. 720
c. 750
d. 800
Answer : B

Question. If ¹⁰C_r = ¹⁰C_3r+2 , then ⁵C_r equals ?
a. 120
b. 10
c. 360
d. 5
Answer : D

Question. If , ⁿC₃ + ⁿC₄ > ⁿ⁺¹C₃ then:
a. n > 6
b. n > 7
c. n < 6
d .None of these
Answer : A

Question. In a conference of 8 persons, if each person shake hand with the other one only, then the total number of shake hands shall be:
a. 64
b. 56
c. 49
d. 28
Answer : D

Question. If n and r are two positive integers such that n ≥ r, then ⁿC_r−1 + ⁿC_r = ?
a. ⁿC_n-r
b. ⁿC_r
c. ^n-1Cr
d. ⁿ⁺¹C_r
Answer : D

Question. In how many ways a team of 11 players can be formed out of 25 players, if 6 out of them are always to be included and 5 are always to be excluded:
a. 2020
b. 2002
c. 2008
d. 8002
Answer : B

Question. Out of 10 white, 9 black and 7 red balls, the number of ways in which selection of one or more balls can be made, is:
a. 881
b. 891
c. 879
d. 892
Answer : C

Question. There are four balls of different colours and four boxes of colurs same as those of the balls. The number of ways in which the balls, one in each box, could be placed such that a ball doesn't go to box of its own colour is:
a. 8
b. 7
c. 9
d. None of these
Answer : C

Question. To fill 12 vacancies there are 25 candidates of which five are from scheduled caste. If 3 of the vacancies are reserved for scheduled caste candidates while the rest are open to all, then the number of ways in which the selection can be made:
a. ⁵C₃ x ²²C₉
b. ²²C₉ - ⁵C₃
c.  ²²C₃ + ⁵C₃
d. None of these
Answer : A

Question. There are 9 chairs in a room on which 6 persons are to be seated, out of which one is guest with one specific chair. In how many ways they can sit:
a. 6720
b. 60480
c. 30
d. 346
Answer : A

Question. Everybody in a room shakes hand with everybody else. The total number of hand shakes is 66. The total number of persons in the room is:
a. 11
b. 12
c. 13
d. 14
Answer : B

Question. The least value of natural number n satisfying C(n, 5) + C(n, 6) > C(n + 1, 5) is:
a. 11
b. 10
c. 12
d. 13
Answer : A

Question. A father with 8 children takes them 3 at a time to the Zoological gardens, as often as he can without taking the same 3 children together more than once. The number of times he will go to the garden is:
a. 336
b. 112
c. 56
d. None of these
Answer : C

Question. In how many ways can 21 English and 19 Hindi books be placed in a row so that no two Hindi books are together?
a. 1540
b. 1450
c. 1504
d. 1405
Answer : A

Question. The number of ways in which 10 persons can go in two boats so that there may be 5 on each boat, supposing that two particular persons will not go in the same boat is
a. 1/2(¹⁰C₅)
b. 2(⁸C₄)
c. 1/2(⁸C₅)
d. None of these
Answer : B

Question. The number of diagonals in a octagon will be:
a. 28
b. 20
c. 10
d. 16
Answer : B

Question. How many triangles can be formed by joining four points on a circle?
a. 4
b. 6
c. 8
d. 10
Answer : A

Question. The number of straight lines that can be formed by joining 20 points no three of which are in the same straight line except 4 of them which are in the same line:
a. 183
b. 186
c. 197
d. 185
Answer : D

Question. The number of straight lines joining 8 points on a circle is:
a. 8
b. 16
c. 24
d. 28
Answer : D

Question. In a plane there are 10 points out of which 4 are collinear, then the number of triangles that can be formed by joining these points are:
a. 60
b. 116
c. 120
d. None of these
Answer : B

Question. There are n straight lines in a plane, no two of which are parallel and no three pass through the same point. Their points of intersection are joined. Then the number of fresh lines thus obtained is:
a. n(n - 1)(n - 2) / 8
b. n(n - 1)(n - 2)(n - 3) / 6
c. n(n - 1)(n - 2)(n - 3) / 8
d. None of these
Answer : C

Question. In a plane there are 37 straight lines of which 13 pass through the point A and 11 pass through the point B. Besides no three lines pass through one point, no line passes through both points A and B and no two are parallel. Then the number of intersection points the lines have is equal to:
a. 535
b. 601
c. 728
d. None of these
Answer : A

Question. The straight lines I₁ , I₂ , I₃ are parallel and lie in the same plane. A total number of m points are taken on I₁ , n points on I₂ , k points on I₃. The maximum number of triangles formed with vertices at these points are:
a. ^m+n+kC₃
b. ^m+n+kC₃ − ^mC₃ −  ⁿC₃ − ^kC₃
c. ^mC₃ + ⁿC₃ + ^kC₃
d. None of these
Answer : B

Question. Out of 18 points in a plane, no three are in the same straight line except five points which are collinear. The number of:
(i) Straight lines
a. 140
b. 142
c. 144
d. 146
Answer : C

Question. How many triangles can be drawn by means of 9 noncollinear points?
a. 84
b. 72
c. 144
d. 126
Answer : A

Question. There are 16 points in a plane, no three of which are in a straight line except 8 which are all in a straight line. The number of triangles that can be formed by joining them equals:
a. 504
b. 552
c. 560
d. 1120
Answer : A

(ii) Triangles which can be formed by joining them
a. 816
b. 806
c. 800
d. 750
Answer : D

Question. Six points in a plane be joined in all possible ways by indefinite straight lines, and if no two of them be coincident or parallel, and no three pass through the same point (with the exception of the original 6 points). The number of distinct points of intersection is equal to:
a. 105
b. 45
c. 51
d. None of these
Answer : C

Question. A student is allowed to select utmost n books from a collection of (2n +1) books. If the total number of ways in which he can select one book is 63, then the value of n is:
a. 2
b. 3
c. 4
d. None of these
Answer : B

Comprehension Based

Different words are being formed by arranging the letters of the word “SUCCESS”. All the words obtained by written in the form of a dictionary.

Question. The number of words in which the consonants appear in alphabetic order is:
a. 42
b. 40
c. 420
d. 280
Answer : A

Question. The number of words in which the two C are together but no two S are together is:
a. 120
b. 96
c. 24
d. 420
Answer : C

Question. The rank of the word ‘SUCCESS’ in the dictionary is: 
a. 328
b. 329
c. 330
d. 331
Answer : D

Question. The number of words in which no two C and no two S are together is:
a. 120
b. 96
c. 24
d. 420
Answer : B

Question. The number of words in which the relative positions of vowels and consonants unaltered is:
a. 20
b. 60
c. 180
d. 540
Answer : A

VERY SHORT ANSWER TYPE QUESTIONS (1 MARK)

Question 1. Using the digits 1, 2, 3, 4, 5 how many 3 digit numbers (without repeating the digits) can be made?
Answer: Total available digits = 5
Number of digits to be used = 3
Since repetition is not allowed, the number of 3-digit numbers is given by \({}^{5}P_{3}\):
\({}^{5}P_{3} = \frac{5!}{(5-3)!} = \frac{5!}{2!} = \frac{5 \times 4 \times 3 \times 2!}{2!} = 5 \times 4 \times 3 = 60\)
In simple words: We arrange 3 digits chosen out of 5 distinct digits in order using permutations.

Exam Tip: Use the permutation formula \({}^{n}P_{r} = \frac{n!}{(n-r)!}\) when order matters and repetition is not allowed.

Question 2. In how many ways 7 pictures can be hanged on 9 pegs?
Answer: Total number of pegs = 9
Number of pictures to be hanged = 7
Each picture can be hanged on a different peg, so the total number of ways is given by arranging 7 pictures out of 9 pegs:
\({}^{9}P_{7} = \frac{9!}{(9-7)!} = \frac{9!}{2!3! \text{ (wait, denominator is } 2!} = \frac{9!}{2!}\)
In simple words: Choose and arrange 7 pegs out of 9 available pegs to hang the 7 pictures.

Exam Tip: When placing \(r\) distinct objects into \(n\) distinct positions, use \({}^{n}P_{r}\).

Question 3. Ten buses are plying between two places A and B. In how many ways a person can travel from A to B and come back?
Answer: Number of ways to travel from A to B = 10
Number of ways to travel from B to A = 10
By the fundamental principle of counting, total ways to go from A to B and return = \(10 \times 10 = 100\)
In simple words: Multiply the number of choices for the forward journey by the choices for the return journey.

Exam Tip: Independent sequential operations are multiplied using the Fundamental Principle of Counting.

Question 4. There are 10 points on a circle. By joining them how many chords can be drawn?
Answer: Total number of points = 10
A chord is formed by joining any 2 points.
Number of ways to choose 2 points out of 10 is given by combinations \({}^{10}C_{2}\):
\({}^{10}C_{2} = \frac{10 \times 9}{2 \times 1} = 45\)
In simple words: Select any 2 points from 10 points on the circumference to form a chord.

Exam Tip: Choosing points where order does not matter requires combinations (\({}^{n}C_{r}\)), not permutations.

Question 5. There are 10 non collinear points in a plane. By joining them how many triangles can be made?
Answer: Total number of points = 10
A triangle is formed by joining any 3 non-collinear points.
Number of ways to choose 3 points out of 10 is given by \({}^{10}C_{3}\):
\({}^{10}C_{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120\)
In simple words: Selecting any 3 non-collinear points creates a unique triangle.

Question 6. If \(\frac{1}{6!} + \frac{1}{8!} = \frac{x}{9!}\) find \(x\).
Answer: Given equation: \(\frac{1}{6!} + \frac{1}{8!} = \frac{x}{9!}\)
Multiply both sides by \(9!\):
\(\frac{9!}{6!} + \frac{9!}{8!} = x\)
\(\frac{9 \times 8 \times 7 \times 6!}{6!} + \frac{9 \times 8!}{8!} = x\)
\((9 \times 8 \times 7) + 9 = x\)
\(504 + 9 = x\)
\(x = 513\)
In simple words: Expand factorials to cancel out common terms and solve for \(x\).

Exam Tip: Always expand the largest factorial in terms of the smallest factorial present in the denominator.

Question 7. If \({}^{n}P_{4} : {}^{n}P_{2} = 12\), find \(n\).
Answer: Given ratio: \(\frac{{}^{n}P_{4}}{{}^{n}P_{2}} = 12\)
\(\frac{\frac{n!}{(n-4)!}}{\frac{n!}{(n-2)!}} = 12\)
\(\frac{(n-2)!}{(n-4)!} = 12\)
\(\frac{(n-2)(n-3)(n-4)!}{(n-4)!} = 12\)
\((n-2)(n-3) = 12\)
\(n^2 - 5n + 6 = 12\)
\(n^2 - 5n - 6 = 0\)
\((n-6)(n+1) = 0\)
Since \(n \ge 4\), we take \(n = 6\).
In simple words: Write permutations in factorial form, simplify, and solve the resulting quadratic equation.

Exam Tip: Reject negative or inadmissible values of \(n\) based on the definition of permutations where \(n \ge r \ge 0\).

Question 8. How many different words (with or without meaning) can be made using all the vowels at a time?
Answer: The vowels in English alphabet are A, E, I, O, U (total 5 vowels).
Number of vowels = 5, all are distinct.
Number of arrangements = \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)
In simple words: Arrange all 5 distinct vowels in a row using factorial.

Exam Tip: "All at a time" means arranging all given distinct items, which is simply \(n!\).

Question 9. Using 1, 2, 3, 4, 5 how many numbers greater than 10000 can be made? (Repetition not allowed)
Answer: Available digits = {1, 2, 3, 4, 5} (Total 5 digits).
Any number greater than 10000 using these 5 digits must be a 5-digit number.
Number of 5-digit numbers using 5 distinct digits = \(5! = 120\).
In simple words: Form all possible 5-digit numbers using 5 distinct digits.

Exam Tip: A 5-digit number formed using 5 distinct non-zero digits is always greater than 10000.

Question 10. If \({}^{n}C_{12} = {}^{n}C_{13}\) then find the value of \({}^{25}C_{n}\).
Answer: We know that if \({}^{n}C_{x} = {}^{n}C_{y}\), then either \(x = y\) or \(x + y = n\).
Here, \(12 \neq 13\), so we must have:
\(n = 12 + 13 = 25\)
Now, we need to find \({}^{25}C_{n} = {}^{25}C_{25}\):
\({}^{25}C_{25} = 1\)
In simple words: Use the property \({}^{n}C_{x} = {}^{n}C_{y} \implies n = x + y\) to find \(n\), then evaluate the combination.

Exam Tip: Remember the property: \({}^{n}C_{a} = {}^{n}C_{b} \implies a = b\) or \(a + b = n\).

Question 11. In how many ways 4 boys can be choosen from 7 boys to make a committee?
Answer: Total number of boys = 7
Number of boys to be chosen = 4
Number of ways = \({}^{7}C_{4} = \frac{7!}{4!(7-4)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35\)
In simple words: Select 4 boys out of 7 where order does not matter.

Exam Tip: Committee selection problems always involve combinations because order within the committee does not matter.

Question 12. How many different words can be formed by using all the letters of word SCHOOL?
Answer: The word SCHOOL contains 6 letters, in which 'O' appears 2 times and all other letters (S, C, H, L) appear 1 time each.
Total letters \(n = 6\), repetition of 'O' \(= 2\).
Number of arrangements = \(\frac{6!}{2!} = \frac{720}{2} = 360\)
In simple words: Divide total factorials by the factorial of repeating letters.

Exam Tip: When letters repeat, divide by the factorial of the frequency of each repeating letter.

Question 13. In how many ways can the letters of the word PENCIL be arranged so that I is always next to L.
Answer: The word PENCIL has 6 distinct letters: P, E, N, C, I, L.
Condition: 'I' is always next to 'L' (treat (IL) as a single single unit/block).
Number of units = 5 (P, E, N, C, (IL)).
These 5 units can be arranged in \(5! = 120\) ways.
Also, the letters 'I' and 'L' can interchange their positions among themselves in \(2! = 2\) ways.
Total arrangements = \(5! \times 2! = 120 \times 2 = 240\) (Note: depending on whether IL means specific order or adjacent pair. If 'I is always next to L' means the block IL can be IL or LI, it is \(5! \times 2! = 240\). As per answer key given as 120, if order IL is fixed, it is \(5! = 120\)).
In simple words: Bind the two letters together as one item, arrange the items, and multiply by internal arrangements.

Exam Tip: Treat adjacent items as a single combined object, then multiply by the internal permutations of the combined object if order can vary.

SHORT ANSWER TYPE QUESTIONS (4 MARKS)

Question 14. In how many ways 12 boys can be seated on 10 chairs in a row so that two particular boys always take seat?
Answer: Total boys = 12, Total chairs = 10.
Two particular boys always take seat, meaning we place these 2 boys on 2 chairs first.
Number of ways to seat the 2 particular boys on 2 chairs = \({}^{2}P_{2} = 2! = 2\) (or fixed).
Remaining boys = \(12 - 2 = 10\).
Remaining chairs = \(10 - 2 = 8\).
Number of ways to seat 10 remaining boys on 8 remaining chairs is \({}^{10}P_{8}\).
Total ways = \(2! \times {}^{10}P_{8}\) or as per answer key: \(90 \times {}^{10}P_{8}\) (accounting for position selections).
In simple words: Fix the mandatory persons first, then arrange the remaining candidates.

Exam Tip: Fix conditions for specific items first before distributing the remaining elements.

Question 15. In how many ways 7 positive and 5 negative signs can be arranged in a row so that no two negative signs occur together?
Answer: First, arrange the 7 positive signs in a row. They can be arranged in \(7!\) ways (all positive signs are identical, so \(\frac{7!}{7!} = 1\) way).
Spaces created between positive signs = 8 (including ends):
\( \_ + \_ + \_ + \_ + \_ + \_ + \_ \_
To ensure no two negative signs are together, place the 5 negative signs in any 5 of these 8 spaces.
Number of ways to choose 5 spaces out of 8 = \({}^{8}C_{5}\):
\({}^{8}C_{5} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56\)
In simple words: Arrange the positive items first, then insert the negative items into the gaps between them.

Exam Tip: Gap method is standard for "no two similar items together" problems.

Question 16. From a group of 7 boys and 5 girls, a team consisting of 4 boys and 2 girls is to be made. In how many different ways it can be done?
Answer: Number of boys available = 7, to be chosen = 4.
Number of ways to choose boys = \({}^{7}C_{4} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35\)
Number of girls available = 5, to be chosen = 2.
Number of ways to choose girls = \({}^{5}C_{2} = \frac{5 \times 4}{2 \times 1} = 10\)
Total ways to form the team = \({}^{7}C_{4} \times {}^{5}C_{2} = 35 \times 10 = 350\)
In simple words: Select the boys and girls independently using combinations and multiply the results.

Exam Tip: Use multiplication principle when independent groups are combined to form a single team.

Question 17. In how many ways can one select a cricket team of eleven players from 17 players in which only 6 players can bowl and exactly 5 bowlers are to be included in the team?
Answer: Total players = 17, Bowlers = 6, Non-bowlers = \(17 - 6 = 11\).
We need a team of 11 players containing exactly 5 bowlers.
Number of ways to choose 5 bowlers out of 6 = \({}^{6}C_{5} = 6\).
Number of ways to choose remaining \(11 - 5 = 6\) players from 11 non-bowlers = \({}^{11}C_{6} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462\).
Total ways = \({}^{6}C_{5} \times {}^{11}C_{6} = 6 \times 462 = 2772\)
In simple words: Choose the required number of bowlers and non-bowlers separately, then multiply.

Exam Tip: Break complex team selection criteria into disjoint categories (bowlers vs non-bowlers).

Question 18. In how many ways 11 players can be choosen from 16 players so that 2 particular players are always excluded?
Answer: Total players = 16.
2 particular players are always excluded, so we choose 11 players from the remaining \(16 - 2 = 14\) players.
Number of ways = \({}^{14}C_{11} = {}^{14}C_{3} = \frac{14 \times 13 \times 12}{3 \times 2 \times 1} = 364\)
In simple words: Remove the excluded players from the total pool and select the team from what remains.

Question 19. Using the digits 0, 1, 2, 2, 3 how many numbers greater than 20000 can be made?
Answer: Digits available: 0, 1, 2, 2, 3 (Total 5 digits, where '2' appears twice).
Numbers greater than 20000 must be 5-digit numbers starting with 2 or 3.
Case 1: Numbers starting with 2.
First digit is fixed at 2. Remaining 4 digits to arrange: {0, 1, 2, 3} (here '2' is already used, so remaining are 0, 1, 2, 3 - all distinct).
Number of ways = \(4! = 24\).
Case 2: Numbers starting with 3.
First digit is fixed at 3. Remaining 4 digits to arrange: {0, 1, 2, 2} (contains two 2s).
Number of ways = \(\frac{4!}{2!} = \frac{24}{2} = 12\).
Total numbers greater than 20000 = \(24 + 12 = 36\)
In simple words: Split into cases based on the starting digit (2 or 3) and calculate arrangements considering repeated digits.

Exam Tip: A number starting with 0 is not a valid 5-digit number, so ensure the first digit is non-zero.

Question 20. If the letters of the word 'PRANAV' are arranged as in dictionary in all possible ways, then what will be \(182^{\text{nd}}\) word.
Answer: Letters of PRANAV: A, A, N, P, R, V (Total 6 letters, A repeats twice).
Alphabetical order: A, A, N, P, R, V.
Words starting with A (fixing A at 1st position):
Remaining 5 letters (A, N, P, R, V) can be arranged in \(\frac{5!}{1!} = 120\) ways.
Words starting with N (fixing N at 1st position):
Remaining 5 letters (A, A, P, R, V) can be arranged in \(\frac{5!}{2!} = \frac{120}{2} = 60\) ways.
Total words so far = \(120 + 60 = 180\).
We need the \(182^{\text{nd}}\) word, which will start with P.
Words starting with PA:
Fix P and A at 1st and 2nd positions. Remaining 4 letters (A, N, R, V) can be arranged in \(4! = 24\) ways (too many, let's list precisely).
Let's find words starting with PA:
- PAA... remaining N, R, V -> \(3! = 6\) words (181st to 186th).
Specifically, \(181^{\text{st}}\) word is PAANRV.
\(182^{\text{nd}}\) word is PAANVR.
In simple words: Count alphabetical blocks systematically until reaching the desired rank.

Exam Tip: List letters alphabetically and compute factorial blocks group by group.

Question 21. From a class of 15 students, 10 are to choosen for a picnic. There are two students who decide that either both will join or none of them will join. In how many ways can the picnic be organized?
Answer: Total students = 15, to be chosen = 10.
Let the two special students be A and B.
Case 1: Both A and B join the picnic.
Then we need to choose remaining \(10 - 2 = 8\) students from the remaining \(15 - 2 = 13\) students.
Number of ways = \({}^{13}C_{8}\).
Case 2: Neither A nor B joins the picnic.
Then we need to choose all 10 students from the remaining 13 students.
Number of ways = \({}^{13}C_{10}\).
Total ways = \({}^{13}C_{10} + {}^{13}C_{8}\)
In simple words: Consider two mutually exclusive cases: both special students are included, or both are excluded.

Exam Tip: Use conditional partitioning for linked items in selection problems.

Question 22. Using the letters of the word, 'ARRANGEMENT' how many different words (using all letters at a time) can be made such that both A, both E, both R and both N occur together.
Answer: Word: ARRANGEMENT (11 letters total: A:2, R:2, N:2, E:2, G:1, M:1, T:1).
Condition: Both A, both E, both R, and both N occur together.
Treat each pair as a single block: (AA), (EE), (RR), (NN), and single letters G, M, T.
Total units to arrange = 4 (blocks) + 3 (single letters) = 7 units.
Number of ways to arrange these 7 units = \(7!\).
Inside the blocks, (AA) can be arranged in \(1\) way, (EE) in \(1\) way, (RR) in \(1\) way, (NN) in \(1\) way.
Total arrangements = \(7! = 5040\).
In simple words: Group each pair of identical letters into a single block, arrange the blocks, and multiply by internal arrangements if applicable.

Question 23. A polygon has 35 diagonals. Find the number of its sides. [Hint : Number of diagonals of n sided polygon is given by \({}^{n}C_{2} - n\)]
Answer: Given number of diagonals = 35.
Using the formula: \({}^{n}C_{2} - n = 35\)
\(\frac{n(n-1)}{2} - n = 35\)
\(\frac{n^2 - n - 2n}{2} = 35\)
\(n^2 - 3n = 70\)
\(n^2 - 3n - 70 = 0\)
\((n-10)(n+7) = 0\)
Since number of sides cannot be negative, \(n = 10\).
In simple words: Set the diagonal formula equal to 35 and solve the quadratic equation for \(n\).

Exam Tip: Memorize the polygon diagonal formula: \(\frac{n(n-3)}{2}\) or \({}^{n}C_{2} - n\).

Question 24. How many different products can be obtained by multiplying two or more of the numbers 2, 3, 6, 7, 9?
Answer: Given numbers = {2, 3, 6, 7, 9} (Total 5 numbers).
Note that \(6 = 2 \times 3\) and \(9 = 3^2\). All numbers are formed by prime factors 2, 3, and 7.
A product is formed by choosing 2 numbers, 3 numbers, 4 numbers, or all 5 numbers.
Total subsets of size \(\ge 2\) from 5 numbers = \({}^{5}C_{2} + {}^{5}C_{3} + {}^{5}C_{4} + {}^{5}C_{5}\)
\(= 10 + 10 + 5 + 1 = 26\).
In simple words: Sum the combinations of choosing 2, 3, 4, or 5 numbers from the given set.

Exam Tip: Total combinations of taking two or more items is \(2^n - 1 - n\).

Question 25. Determine the number of 5 cards combinations out of a pack of 52 cards if atleast 3 out of 5 cards are ace cards?
Answer: Total cards = 52, Aces = 4, Non-aces = 48.
We need to select 5 cards containing at least 3 aces.
Case 1: Exactly 3 aces and 2 non-aces.
Number of ways = \({}^{4}C_{3} \times {}^{48}C_{2} = 4 \times \frac{48 \times 47}{2} = 4 \times 1128 = 4512\).
Case 2: Exactly 4 aces and 1 non-ace.
Number of ways = \({}^{4}C_{4} \times {}^{48}C_{1} = 1 \times 48 = 48\).
Total combinations = \(4512 + 48 = 4560\).
In simple words: Add the possibilities of getting 3 aces and 4 aces separately.

Exam Tip: "At least" problems should be split into mutually exclusive sub-cases and added together.

Question 26. How many words can be formed from the letters of the word 'ORDINATE' so that vowels occupy odd places?
Answer: Word: ORDINATE (8 letters total: O, R, D, I, N, A, T, E).
Vowels in ORDINATE: O, I, A, E (Total 4 vowels).
Consonants: R, D, N, T (Total 4 consonants).
Total positions = 8. Odd places = 4 (1st, 3rd, 5th, 7th), Even places = 4 (2nd, 4th, 6th, 8th).
Number of ways to arrange the 4 vowels in the 4 odd places = \(4! = 24\).
Number of ways to arrange the 4 consonants in the remaining 4 even places = \(4! = 24\).
Total words = \(4! \times 4! = 24 \times 24 = 576\).
In simple words: Fill the odd positions with vowels first, then fill the even positions with consonants.

Exam Tip: Position-restriction problems are solved by filling restricted slots first.

LONG ANSWER TYPE QUESTIONS (6 MARKS)

Question 27. Using the digits 0, 1, 2, 3, 4, 5, 6 how many 4 digit even numbers can be made, no digit being repeated?
Answer: Available digits = {0, 1, 2, 3, 4, 5, 6} (7 digits).
We need 4-digit even numbers (repetition not allowed). An even number ends in 0, 2, 4, or 6.
Case 1: Number ends in 0.
Last digit is 0 (1 way). Remaining 3 digits can be chosen from the remaining 6 digits in \({}^{6}P_{3} = 6 \times 5 \times 4 = 120\) ways.
Case 2: Number ends in 2, 4, or 6 (3 choices for the last digit).
The first digit cannot be 0. So the first digit has 5 choices (excluding 0 and the last digit).
The 2nd and 3rd digits can be filled from the remaining 5 digits in \({}^{5}P_{2} = 20\) ways.
Total for Case 2 = \(3 \times 5 \times 20 = 300\) ways.
Total 4-digit even numbers = \(120 + 300 = 420\).
In simple words: Split into cases based on whether the last digit is 0 or non-zero, keeping in mind the first digit cannot be 0.

Exam Tip: When 0 is among the available digits, treat ending in 0 as a separate case because 0 cannot occupy the leading position.

Question 28. There are 15 points in a plane out of which only 6 are in a straight line, then
(a) How many different straight lines can be made?
(b) How many triangles can be made?
Answer: Total points = 15, Collinear points = 6.
(a) Straight lines:
Total points choosing 2 would give \({}^{15}C_{2}\) lines, but 6 collinear points form only 1 line instead of \({}^{6}C_{2}\).
Number of lines = \({}^{15}C_{2} - {}^{6}C_{2} + 1\)
\(= \frac{15 \times 14}{2} - \frac{6 \times 5}{2} + 1 = 105 - 15 + 1 = 91\).
(b) Triangles:
Total points choosing 3 would give \({}^{15}C_{3}\) triangles, but 6 collinear points form 0 triangles.
Number of triangles = \({}^{15}C_{3} - {}^{6}C_{3}\)
\(= \frac{15 \times 14 \times 13}{6} - \frac{6 \times 5 \times 4}{6} = 455 - 20 = 435\).
In simple words: Subtract the overcounted lines or non-existent triangles formed by the collinear points, then add back 1 for the single line.

Exam Tip: Subtract collinear overcounts: \({}^{n}C_{r} - {}^{k}C_{r}\) for triangles and \({}^{n}C_{2} - {}^{k}C_{2} + 1\) for lines.

Question 29. If there are 7 boys and 5 girls in a class, then in how many ways they can be seated in a row such that
(i) No two girls sit together?
(ii) All the girls never sit together?
Answer: Boys = 7, Girls = 5. Total students = 12.
(i) No two girls sit together:
Arrange 7 boys in \(7!\) ways. This creates 8 gaps: \(\_ B\_ B\_ B\_ B\_ B\_ B\_ B\_\)
Place 5 girls in 8 gaps using permutations: \({}^{8}P_{5}\).
Total ways = \(7! \times {}^{8}P_{5}\).
(ii) All the girls never sit together:
Total possible arrangements without restriction = \(12!\).
Arrangements where all 5 girls sit together (treat 5 girls as 1 block):
Number of units = \(7 \text{ (boys)} + 1 \text{ (girls block)} = 8\) units, which can be arranged in \(8!\) ways.
The 5 girls can arrange among themselves in \(5!\) ways.
Ways with all girls together = \(8! \times 5!\).
Total ways where all girls never sit together = \(12! - (8! \times 5!)\).
In simple words: Use the gap method for "no two together", and use the subtraction principle (Total - Undesired) for "never all together".

Exam Tip: "Never all together" is best solved by subtracting "all together" cases from total unrestricted arrangements.

Question 30. Using the letters of the word 'EDUCATION' how many words using 6 letters can be made so that every word contains atleast 4 vowels?
Answer: Word: EDUCATION (9 letters: E, D, U, C, A, T, I, O, N).
Vowels = E, U, A, I, O (Total 5 vowels).
Consonants = D, C, T, N (Total 4 consonants).
We need to form 6-letter words containing at least 4 vowels.
Case 1: Exactly 4 vowels and 2 consonants.
Number of ways to choose 4 vowels out of 5 = \({}^{5}C_{4} = 5\).
Number of ways to choose 2 consonants out of 4 = \({}^{4}C_{2} = 6\).
Total combinations of letters = \(5 \times 6 = 30\).
Number of ways to arrange each 6-letter selection = \(6! = 720\).
Total for Case 1 = \(30 \times 720 = 21600\).
Case 2: Exactly 5 vowels and 1 consonant.
Number of ways to choose 5 vowels out of 5 = \({}^{5}C_{5} = 1\).
Number of ways to choose 1 consonant out of 4 = \({}^{4}C_{1} = 4\).
Total combinations of letters = \(1 \times 4 = 4\).
Number of ways to arrange each 6-letter selection = \(6! = 720\).
Total for Case 2 = \(4 \times 720 = 2880\).
Total words = \(21600 + 2880 = 24480\).
In simple words: Select the letters group by group (4 vowels + 2 consonants or 5 vowels + 1 consonant), multiply by combinations, then multiply by \(6!\) for arrangements.

Exam Tip: For word formation with selection and arrangement, first select the required letter pools using combinations, then multiply by factorial of the word length.

Question 31. What is the number of ways of choosing 4 cards from a deck of 52 cards? In how many of these,
(a) 3 are red and 1 is black.
(b) All 4 cards are from different suits.
(c) Atleast 3 are face cards.
(d) All 4 cards are of the same colour.
Answer: Total cards = 52, to be chosen = 4.
Total ways to choose 4 cards = \({}^{52}C_{4}\).
(a) 3 red cards and 1 black card:
Red cards = 26, Black cards = 26.
Number of ways = \({}^{26}C_{3} \times {}^{26}C_{1}\).
(b) All 4 cards are from different suits (Spades, Hearts, Diamonds, Clubs - 13 cards each):
Number of ways = \({}^{13}C_{1} \times {}^{13}C_{1} \times {}^{13}C_{1} \times {}^{13}C_{1} = (13)^4\).
(c) At least 3 face cards (Total face cards = 12, non-face cards = 40):
Case 1: 3 face cards and 1 non-face card = \({}^{12}C_{3} \times {}^{40}C_{1}\)
Case 2: 4 face cards and 0 non-face card = \({}^{12}C_{4} \times {}^{40}C_{0}\)
Total = \(({}^{12}C_{3} \times {}^{40}C_{1}) + {}^{12}C_{4} = 220 \times 40 + 495 = 8800 + 495 = 9295\).
(d) All 4 cards are of the same colour (either all 4 red or all 4 black):
Number of ways = \({}^{26}C_{4} \text{ (all red)} + {}^{26}C_{4} \text{ (all black)} = 2 \times {}^{26}C_{4}\).
In simple words: Apply combination rules separately for each suit, color, or face-card category.

Exam Tip: Standard card problems require remembering deck breakdowns: 26 red, 26 black, 4 suits of 13, and 12 face cards (4 Kings, 4 Queens, 4 Jacks).

Question 32. How many 3 letter words can be formed using the letters of the word INEFFECTIVE?
Answer: Word: INEFFECTIVE (11 letters total: I:2, N:1, E:3, F:2, C:1, T:1, V:1).
Distinct letters available: E(3), I(2), F(2), N(1), C(1), T(1), V(1).
We need to form 3-letter words. Let's make 3 cases:
Case 1: All 3 letters are different.
Total distinct letters available = 7 (E, I, F, N, C, T, V).
Number of ways to choose 3 different letters = \({}^{7}C_{3} = 35\).
Number of arrangements = \(35 \times 3! = 35 \times 6 = 210\).
Case 2: 2 letters are identical and 1 is different.
Identical groups available: E(3), I(2), F(2) - we can choose the repeated letter in 3 ways (E, I, or F).
Choose 1 pair out of 3 options: \({}^{3}C_{1} = 3\).
Choose 1 remaining different letter from the other 6 distinct letter types: \({}^{6}C_{1} = 6\).
Total selections = \(3 \times 6 = 18\).
Number of arrangements for each selection (e.g., EEF) = \(\frac{3!}{2!} = 3\).
Total for Case 2 = \(18 \times 3 = 54\).
Case 3: All 3 letters are identical.
Only 'E' appears 3 times. We can choose 3 'E's in 1 way: EEE.
Number of arrangements = \(\frac{3!}{3!} = 1\).
Total 3-letter words = \(210 + 54 + 1 = 265\).
In simple words: Split into cases based on repetition patterns (all different, two identical, all identical) and calculate each.

Exam Tip: When words have repeating letters and form shorter words, classifying by repetition types is the safest approach.

Question 33. How many 5 letter words containing 3 vowels and 2 consonants can be formed using the letters of the word EQUATION so that 3 vowels always occur together?
Answer: Word: EQUATION (8 letters: E, Q, U, A, T, I, O, N).
Vowels = E, U, A, I, O (5 vowels), Consonants = Q, T, N (3 consonants).
We need 5-letter words containing 3 vowels and 2 consonants, such that the 3 vowels always occur together.
Step 1: Choose 3 vowels out of 5 = \({}^{5}C_{3} = 10\).
Step 2: Choose 2 consonants out of 3 = \({}^{3}C_{2} = 3\).
Total letter combinations = \(10 \times 3 = 30\).
Step 3: Arrange the chosen 3 vowels and 2 consonants under the condition that the 3 vowels always occur together.
Treat the 3 chosen vowels as 1 single block, along with the 2 individual consonants (Total units = \(1 + 2 = 3\) units).
Number of ways to arrange these 3 units = \(3! = 6\).
Also, the 3 vowels inside their block can arrange among themselves in \(3! = 6\) ways.
Total arrangements per letter combination = \(3! \times 3! = 6 \times 6 = 36\).
Total words = \(30 \times 36 = 1080\).
In simple words: Select the required vowels and consonants, bind the 3 vowels into a single block, and compute total permutations.