CBSE Class 11 Mathematics Permutations and Combinations Notes Set 02

Important Notes and Formulas

The Factorial

Factorial notation: Let n be a positive integer. Then, the continued product of first n natural numbers is called factorial n, to be denoted by n! Also, we define 0! = 1.
When n is negative or a fraction, n! is not defined.
Thus, n! = n(n − 1)(n − 2) …… 3.2.1.
Deduction: n! = n(n − 1)(n − 2)(n − 3) …… 3.2.1 = n[(n − 1)(n − 2)(n − 3) …… 3.2.1] = n[(n − 1)!]
Thus, 5! = 5 × (4!), 3! = 3 × (2!) and 2! = 2 × (1!)
Also, 1! = 1 × (0!) ⇒ 0! = 1.

Fundamental Principles of Counting

(1) Addition principle: Suppose that A and B are two disjoint events (mutually exclusive); that is, they never occur together. Further suppose that A occurs in m ways and B in n ways. Then A or B can occur in m + n ways. This rule can also be applied to more than two mutually exclusive events.

(2) Multiplication principle: Suppose that an event X can be decomposed into two stages A and B. Let stage A occur in m ways and suppose that these stages are unrelated, in the sense that stage B occurs in n ways regardless of the outcome of stage A. Then event X occurs in mn ways. This rule is applicable even if event X can be decomposed in more than two stages.
Note: The above principle can be extended for any finite number of operations and may be stated as under:
If one operation can be performed independently in m different ways and if second operation can be performed independently in n different ways and a third operation can be performed independently in p different ways and so on, then the total number of ways in which all the operations can be performed in the stated order is (m × n × p × …).

Definition of Permutation

The ways of arranging or selecting a smaller or an equal number of persons or objects at a time from a given group of persons or objects with due regard being paid to the order of arrangement or selection are called the (different) permutations.
For example: Three different things a, b and c are given, then different arrangements which can be made by taking two things from three given things are ab, ac, bc, ba, ca, cb.
Therefore the number of permutations will be 6.

Number of Permutations without Repetition

(1) Arranging n objects, taken r at a time is equivalent to filling r places from n things.
r places: 1st, 2nd, 3rd, 4th, ……, rth
Number of choices: n, (n − 1), (n − 2), (n − 3), ……, (n − r + 1)
The number of ways of arranging = The number of ways of filling r places
= \( n(n-1)(n-2)\ldots(n-r+1) = \frac{n(n-1)(n-2)\ldots(n-r+1)\,(n-r)!}{(n-r)!} = \frac{n!}{(n-r)!} = {}^{n}P_{r} \)

(2) The number of arrangements of n different objects taken all at a time = \( {}^{n}P_{n} = n! \)
Note: \( {}^{n}P_{0} = \frac{n!}{n!} = 1 \); \( {}^{n}P_{r} = n \cdot {}^{n-1}P_{r-1} \); 0! = 1.

Number of Permutations with Repetition

(1) The number of permutations (arrangements) of n different objects, taken r at a time, when each object may occur once, twice, thrice, …… up to r times in any arrangement = The number of ways of filling r places where each place can be filled by any one of n objects.
r places: 1st, 2nd, 3rd, 4th, ……, rth
Number of choices: n, n, n, n, ……, n
The number of permutations = The number of ways of filling r places = n^r.

(2) The number of arrangements that can be formed using n objects out of which p are identical (and of one kind), q are identical (and of another kind), r are identical (and of another kind) and the rest are distinct is \( \frac{n!}{p!\,q!\,r!} \).

Conditional Permutations

(1) Number of permutations of n dissimilar things taken r at a time when p particular things always occur = \( {}^{n-p}C_{r-p}\; r! \)

(2) Number of permutations of n dissimilar things taken r at a time when p particular things never occur = \( {}^{n-p}C_{r}\; r! \)

(3) The total number of permutations of n different things taken not more than r at a time, when each thing may be repeated any number of times, is \( \frac{n(n^{r}-1)}{n-1} \).

(4) Number of permutations of n different things, taken all at a time, when m specified things always come together is m! × (n − m + 1)!

(5) Number of permutations of n different things, taken all at a time, when m specified things never come together is n! − m! × (n − m + 1)!

(6) Let there be n objects, of which m objects are alike of one kind, and the remaining (n − m) objects are alike of another kind. Then, the total number of mutually distinguishable permutations that can be formed from these objects is \( \frac{n!}{m!\times(n-m)!} \).
Note: The above theorem can be extended further, i.e., if there are n objects, of which p₁ are alike of one kind; p₂ are alike of another kind; p₃ are alike of 3rd kind; ……; p_r are alike of r^th kind such that p₁ + p₂ + …… + p_r = n; then the number of permutations of these n objects is \( \frac{n!}{(p_1!)\times(p_2!)\times\ldots\times(p_r!)} \).

Important Tips
Gap method: Suppose 5 males A, B, C, D, E are arranged in a row as × A × B × C × D × E ×. There will be six gaps between these five: four in between and two at either end. Now if three females P, Q, R are to be arranged so that no two are together, we use the gap method, i.e., arrange them in between these 6 gaps. Hence the answer will be \( {}^{6}P_{3} \).
Together: Suppose we have to arrange 5 persons in a row which can be done in 5! = 120 ways. But if two particular persons are to be together always, then we tie these two particular persons with a string. Thus we have 5 − 2 + 1 (1 corresponding to these two together) = 3 + 1 = 4 units, which can be arranged in 4! ways. Now we loosen the string and these two particular persons can be arranged in 2! ways. Thus total arrangements = 24 × 2 = 48.
Never together = Total − Together = 120 − 48 = 72.

Combinations: Definition

Each of the different groups or selections which can be formed by taking some or all of a number of objects, irrespective of their arrangements, is called a combination.
Suppose we want to select two out of three persons A, B and C. We may choose AB or BC or AC. Clearly, AB and BA represent the same selection or group but they give rise to different arrangements. Clearly, in a group or selection, the order in which the objects are arranged is immaterial.
Notation: The number of all combinations of n things, taken r at a time is denoted by C(n, r) or \( {}^{n}C_{r} \) or \( \binom{n}{r} \).

Difference between a permutation and combination:
(i) In a combination only selection is made whereas in a permutation not only a selection is made but also an arrangement in a definite order is considered.
(ii) In a combination, the ordering of the selected objects is immaterial whereas in a permutation, the ordering is essential. For example A, B and B, A are same as combination but different as permutations.
(iii) Practically to find the permutation of n different items, taken r at a time, we first select r items from n items and then arrange them. So usually the number of permutations exceeds the number of combinations.
(iv) Each combination corresponds to many permutations. For example, the six permutations ABC, ACB, BCA, BAC, CBA and CAB correspond to the same combination ABC.
Note: Generally we use the word 'arrangements' for permutations and the word 'selection' for combinations.

Number of Combinations without Repetition

The number of combinations (selections or groups) that can be formed from n different objects taken r (0 ≤ r ≤ n) at a time is \( {}^{n}C_{r} = \frac{n!}{r!\,(n-r)!} \).
Let the total number of selections (or groups) = x. Each group contains r objects, which can be arranged in r! ways. Hence the number of arrangements of r objects = x × (r!). But the number of arrangements = \( {}^{n}P_{r} \).
\( \Rightarrow x \times (r!) = {}^{n}P_{r} \Rightarrow x = \frac{{}^{n}P_{r}}{r!} \Rightarrow x = \frac{n!}{r!\,(n-r)!} = {}^{n}C_{r} \)

Important Tips
(i) \( {}^{n}C_{r} \) is a natural number.
(ii) \( {}^{n}C_{0} = {}^{n}C_{n} = 1,\ {}^{n}C_{1} = n \)
(iii) \( {}^{n}C_{r} = {}^{n}C_{n-r} \)
(iv) \( {}^{n}C_{r} + {}^{n}C_{r-1} = {}^{n+1}C_{r} \)
(v) \( {}^{n}C_{x} = {}^{n}C_{y} \Leftrightarrow x = y \) or x + y = n
(vi) \( n \cdot {}^{n-1}C_{r-1} = (n-r+1)\, {}^{n}C_{r-1} \)
(vii) If n is even, then the greatest value of \( {}^{n}C_{r} \) is \( {}^{n}C_{n/2} \). If n is odd, then the greatest value of \( {}^{n}C_{r} \) is \( {}^{n}C_{\frac{n+1}{2}} \) or \( {}^{n}C_{\frac{n-1}{2}} \).
(viii) \( {}^{n}C_{r} = \frac{n}{r} \cdot {}^{n-1}C_{r-1} \)
(ix) \( \frac{{}^{n}C_{r}}{{}^{n}C_{r-1}} = \frac{n-r+1}{r} \)
(x) \( {}^{n}C_{0} + {}^{n}C_{1} + {}^{n}C_{2} + \ldots + {}^{n}C_{n} = 2^{n} \)
(xi) \( {}^{n}C_{0} + {}^{n}C_{2} + {}^{n}C_{4} + \ldots = {}^{n}C_{1} + {}^{n}C_{3} + {}^{n}C_{5} + \ldots = 2^{n-1} \)
Note: Number of combinations of n dissimilar things taken all at a time is \( {}^{n}C_{n} = \frac{n!}{n!\,(n-n)!} = \frac{1}{0!} = 1 \) (∵ 0! = 1).

All Possible Selections

(1) The total number of ways in which it is possible to form groups by taking some or all of n different things at a time is 2ⁿ − 1.
(2) The number of selections of r objects out of n identical objects is 1.
(3) Total number of selections of zero or more objects from n identical objects is n + 1.

Conditional Combinations

The number of ways in which r objects can be selected from n different objects if k particular objects are
(i) Always included = \( {}^{n-k}C_{r-k} \)
(ii) Never included = \( {}^{n-k}C_{r} \)

Some Important Results for Geometrical Problems

(1) Number of total different straight lines formed by joining the n points on a plane of which m (< n) are collinear is \( {}^{n}C_{2} - {}^{m}C_{2} + 1 \).
(2) Number of total triangles formed by joining the n points on a plane of which m (< n) are collinear is \( {}^{n}C_{3} - {}^{m}C_{3} \).
(3) Number of diagonals in a polygon of n sides is \( {}^{n}C_{2} - n \).
(4) If m parallel lines in a plane are intersected by a family of other n parallel lines, then the total number of parallelograms so formed is \( {}^{m}C_{2} \times {}^{n}C_{2} \), i.e., \( \frac{mn(m-1)(n-1)}{4} \).
(5) Given n points on the circumference of a circle, then
(i) Number of straight lines = \( {}^{n}C_{2} \)   (ii) Number of triangles = \( {}^{n}C_{3} \)   (iii) Number of quadrilaterals = \( {}^{n}C_{4} \).

Divisibility Rules (useful in problems on forming numbers)

(i) All the numbers whose last digit is an even number 0, 2, 4, 6 or 8 are divisible by 2.
(ii) All the numbers the sum of whose digits is divisible by 3 are divisible by 3, e.g. 534: sum of the digits is 12, which is divisible by 3, and hence 534 is also divisible by 3.
(iii) All those numbers whose last two-digit number is divisible by 4 are divisible by 4, e.g. 7312, 8936 are such that 12, 36 are divisible by 4 and hence the given numbers are also divisible by 4.
(iv) All those numbers which have either 0 or 5 as the last digit are divisible by 5.
(v) All those numbers which are divisible by 2 and 3 simultaneously are divisible by 6, e.g. 108, 756 etc.
(vi) All those numbers whose last three-digit number is divisible by 8 are divisible by 8.
(vii) All those numbers the sum of whose digits is divisible by 9 are divisible by 9.
(viii) All those numbers whose last two digits are divisible by 25 are divisible by 25, e.g. 73125, 2400 etc.

Solved Examples

Example 1. 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:
Ans. (c)
The student has seven choices from the morning courses out of which he can select one course in 7 ways.
For the evening course, he has 5 choices out of which he can select one course in 5 ways.
Hence he has total number of 7 + 5 = 12 choices.

Example 2. 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:
Ans. (a)
There are 12 questions in exercise 7. So, one question from exercise 7 can be selected in 12 ways.
Exercise 8 contains 18 questions. So, second question can be selected in 18 ways.
There are 9 questions in exercise 9. So, third question can be selected in 9 ways.
Hence, three questions can be selected in 12 × 18 × 9 = 1944 ways.

Example 3. If \( {}^{n}P_{4} : {}^{n}P_{5} = 1 : 2 \), then n = [MP PET 1987; Rajasthan PET 1996]
(a) 4
(b) 5
(c) 6
(d) 7
Answer:
Ans. (c)
\( \frac{{}^{n}P_{4}}{{}^{n}P_{5}} = \frac{1}{2} \Rightarrow \frac{n!}{(n-4)!} \times \frac{(n-5)!}{n!} = \frac{1}{2} \Rightarrow \frac{1}{n-4} = \frac{1}{2} \)
⇒ n − 4 = 2 ⇒ n = 6.

Example 4. In a train 5 seats are vacant, then in how many ways can three passengers sit [Rajasthan PET 1985; MP PET 2003]
(a) 20
(b) 30
(c) 60
(d) 10
Answer:
Ans. (c)
Number of ways = \( {}^{5}P_{3} = \frac{5!}{(5-3)!} = \frac{5!}{2!} = \frac{120}{2} = 60 \).

Example 5. How many words comprising of any three letters of the word “UNIVERSAL” can be formed
(a) 504
(b) 405
(c) 540
(d) 450
Answer:
Ans. (a)
The word UNIVERSAL has 9 different letters.
Required number of words = \( {}^{9}P_{3} = \frac{9!}{(9-3)!} = \frac{9!}{6!} = 504 \).

Example 6. How many numbers of five digits can be formed from the numbers 2, 0, 4, 3, 8 when repetition of digits is not allowed [MP PET 2000]
(a) 96
(b) 120
(c) 144
(d) 14
Answer:
Ans. (a)
Given numbers are 2, 0, 4, 3, 8.
Numbers that can be formed = {Total − Those beginning with 0} = {5! − 4!} = 120 − 24 = 96.

Example 7. 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:
Ans. (c)
All the 5 digit numbers and 6 digit numbers are greater than 3000. Number of 5 digit numbers = \( {}^{6}P_{5} - {}^{5}P_{4} = 720 - 120 = 600 \) (since the arrangements with 0 at the ten thousand place should be omitted).
Similarly, number of 6 digit numbers = 6! − 5! = 600.
Now the 4 digit numbers which are greater than 3000 have 3, 4 or 5 at the first place; this can be done in 3 ways and the remaining 3 places may be filled from the remaining 5 digits, i.e., required number of 4 digit numbers = \( {}^{5}P_{3} \times 3 = 180 \).
Hence total required number of numbers = 600 + 600 + 180 = 1380.

Example 8. The number of arrangements of the letters of the word “CALCUTTA” is [MP PET 1984]
(a) 2520
(b) 5040
(c) 10080
(d) 40320
Answer:
Ans. (b)
Required number of ways = \( \frac{8!}{2!\,2!\,2!} = 5040 \) [since here there are 2 C's, 2 T's and 2 A's].

Example 9. The number of 5 digit telephone numbers having at least one of their digits repeated is
(a) 90,000
(b) 100,000
(c) 30,240
(d) 69,760
Answer:
Ans. (d)
Using the digits 0, 1, 2, ……, 9 the number of five digit telephone numbers which can be formed is 10⁵ (since repetition is allowed).
The number of five digit telephone numbers which have none of the digits repeated = \( {}^{10}P_{5} = 30240 \).
∴ The required number of telephone numbers = 10⁵ − 30240 = 69760.

Example 10. How many words can be made from the letters of the word ‘COMMITTEE’ [MP PET 2002; RPET 1986]
(a) \( \frac{9!}{(2!)^2} \)
(b) \( \frac{9!}{(2!)^3} \)
(c) \( \frac{9!}{2!} \)
(d) 9!
Answer:
Ans. (b)
Number of words = \( \frac{9!}{2!\,2!\,2!} = \frac{9!}{(2!)^3} \) [since here the total number of letters is 9, with 2 M's, 2 T's and 2 E's].

Example 11. All the letters of the word ‘EAMCET’ are arranged in all possible ways. The number of such arrangements in which no two vowels are adjacent to each other is [EAMCET 1987; DCE 2000]
(a) 360
(b) 114
(c) 72
(d) 54
Answer:
Ans. (c)
First we arrange the 3 consonants M, C, T in 3! ways and then at the four places (two places between them and two places on the two sides) the 3 vowels E, A, E can be placed in \( {}^{4}P_{3} \times \frac{1}{2!} \) ways.
Hence the required number of ways = \( 3! \times {}^{4}P_{3} \times \frac{1}{2!} = 72 \).

Example 12. The number of words which can be made out of the letters of the word ‘MOBILE’ when consonants always occupy odd places is
(a) 20
(b) 36
(c) 30
(d) 720
Answer:
Ans. (b)
The word ‘MOBILE’ has three even places and three odd places. It has 3 consonants and 3 vowels. In the three odd places we have to fix up 3 consonants, which can be done in \( {}^{3}P_{3} \) ways.
The remaining three places can be filled with the three vowels in \( {}^{3}P_{3} \) ways.
The total number of ways = \( {}^{3}P_{3} \times {}^{3}P_{3} = 36 \).

Example 13. The number of 4 digit numbers that can be formed from the digits 0, 1, 2, 3, 4, 5, 6, 7 so that each number contains the digit 1 (repetition of digits not allowed)
(a) 1225
(b) 1252
(c) 1522
(d) 750
Answer:
Ans. (d)
After fixing 1 at one position out of 4 places, the remaining 3 places can be filled in \( {}^{7}P_{3} \) ways. So the number of such arrangements = 4 × \( {}^{7}P_{3} \) = 4 × 210 = 840.
But some of these numbers have 0 in the thousands place. In these, 1 is at one of the remaining 3 places and the other 2 places are filled in \( {}^{6}P_{2} \) ways, i.e., 3 × \( {}^{6}P_{2} \) = 3 × 30 = 90 ways.
∴ Total ways = 840 − 90 = 750.

Example 14. 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) \( \frac{m!\,(m+1)!}{(m-n+1)!} \)
(b) \( \frac{m!\,(m-1)!}{(m-n+1)!} \)
(c) \( \frac{(m-1)!\,(m+1)!}{(m-n+1)!} \)
(d) None of these
Answer:
Ans. (a)
First arrange m men in a row in m! ways. Since n < m and no two women can sit together, in any one of the m! arrangements there are (m + 1) places in which the n women can be arranged in \( {}^{m+1}P_{n} \) ways.
∴ By the fundamental theorem, the required number of arrangements = \( m! \times {}^{m+1}P_{n} = \frac{m!\,(m+1)!}{(m-n+1)!} \).

Example 15. 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:
Ans. (a)
Words starting from A are 5! = 120; words starting from I are 5! = 120;
Words starting from KA are 4! = 24; words starting from KI are 4! = 24;
Words starting from KN are 4! = 24; words starting from KRA are 3! = 6;
Words starting from KRIA are 2! = 2; words starting from KRIN are 2! = 2;
Words starting from KRISA are 1! = 1; and then KRISNA itself is 1.
Hence the rank of the word KRISNA is 120 + 120 + 24 + 24 + 24 + 6 + 2 + 2 + 1 + 1 = 324.

Example 16. We are to form different words with the letters of the word ‘INTEGER’. Let m₁ be the number of words in which I and N are never together, and m₂ be the number of words which begin with I and end with R. Then m₁/m₂ is equal to [AMU 2000]
(a) 30
(b) 60
(c) 90
(d) 180
Answer:
Ans. (a)
We have 5 letters other than ‘I’ and ‘N’ of which two are identical (E's). We can arrange these letters in a line in \( \frac{5!}{2!} \) ways. In any such arrangement ‘I’ and ‘N’ can be placed in the 6 available gaps in \( {}^{6}P_{2} \) ways, so the required number = \( \frac{5!}{2!} \cdot {}^{6}P_{2} = m_1 \).
Now, if the word starts with I and ends with R, then the remaining letters are 5 (with two E's). So, total number of ways = \( \frac{5!}{2!} = m_2 \).
∴ \( \frac{m_1}{m_2} = {}^{6}P_{2} = \frac{6!}{4!} = 30 \).

Example 17. 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 [IIT 1998]
(a) 6
(b) 7
(c) 8
(d) 9
Answer:
Ans. (b)
Since at any place, any of the digits 2, 5 and 7 can be used, the total number of such positive n-digit numbers is 3ⁿ.
Since we have to form 900 distinct numbers, 3ⁿ ≥ 900. As 3⁶ = 729 and 3⁷ = 2187, the smallest value is n = 7.

Example 18. 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:
Ans. (b)
The 4 odd digits 1, 3, 3, 1 can be arranged in the 4 odd places in \( \frac{4!}{2!\,2!} = 6 \) ways.
The 3 even digits 2, 4, 2 can be arranged in the three even places in \( \frac{3!}{2!} = 3 \) ways.
Hence the required number of ways = 6 × 3 = 18.

Example 19. If \( {}^{15}C_{3r} = {}^{15}C_{r+3} \), then the value of r is [IIT 1967; Rajasthan PET 1991; MP PET 1998; Karnataka CET 1996]
(a) 3
(b) 4
(c) 5
(d) 8
Answer:
Ans. (a)
\( {}^{15}C_{3r} = {}^{15}C_{r+3} \Rightarrow {}^{15}C_{15-3r} = {}^{15}C_{r+3} \Rightarrow 15 - 3r = r + 3 \Rightarrow r = 3 \).
(The other possibility 3r = r + 3 gives r = 3/2, which is not an integer.)

Example 20. \( \frac{{}^{n}C_{r}}{{}^{n}C_{r-1}} = \) [MP PET 1984]
(a) \( \frac{n-r}{r} \)
(b) \( \frac{n+r-1}{r} \)
(c) \( \frac{n-r+1}{r} \)
(d) \( \frac{n-r-1}{r} \)
Answer:
Ans. (c)
\( \frac{{}^{n}C_{r}}{{}^{n}C_{r-1}} = \frac{n!}{r!\,(n-r)!} \times \frac{(r-1)!\,(n-r+1)!}{n!} \)
\( = \frac{(n-r+1)\,(r-1)!\,(n-r)!}{r\,(r-1)!\,(n-r)!} = \frac{n-r+1}{r} \).

Example 21. If \( {}^{n+1}C_{3} = 2\,{}^{n}C_{2} \), then n = [MP PET 2000]
(a) 3
(b) 4
(c) 5
(d) 6
Answer:
Ans. (c)
\( {}^{n+1}C_{3} = 2 \cdot {}^{n}C_{2} \Rightarrow \frac{(n+1)!}{3!\,(n-2)!} = 2 \cdot \frac{n!}{2!\,(n-2)!} \)
\( \Rightarrow \frac{n+1}{3 \cdot 2!} = \frac{2}{2!} \Rightarrow n + 1 = 6 \Rightarrow n = 5 \).

Example 22. If \( {}^{n}C_{r-1} = 36,\ {}^{n}C_{r} = 84 \) and \( {}^{n}C_{r+1} = 126 \), then the value of r is [IIT 1979; Pb. CET 1993; DCE 1999; MP PET 2001]
(a) 1
(b) 2
(c) 3
(d) None of these
Answer:
Ans. (c)
Here \( \frac{{}^{n}C_{r-1}}{{}^{n}C_{r}} = \frac{36}{84} \Rightarrow \frac{r}{n-r+1} = \frac{3}{7} \Rightarrow 3n - 10r = -3 \)
and \( \frac{{}^{n}C_{r}}{{}^{n}C_{r+1}} = \frac{84}{126} \Rightarrow \frac{r+1}{n-r} = \frac{2}{3} \Rightarrow 2n - 5r = 3 \), i.e., 4n − 10r = 6.
On solving, we get n = 9 and r = 3.

Example 23. In a conference of 8 persons, if each person shakes hand with the other one only, then the total number of hand shakes shall be
(a) 64
(b) 56
(c) 49
(d) 28
Answer:
Ans. (d)
Total number of hand shakes when each person shakes hand with each other once only = \( {}^{8}C_{2} = 28 \) ways.

Example 24. How many words of 4 consonants and 3 vowels can be formed from 6 consonants and 5 vowels [Rajasthan PET]
(a) 75000
(b) 756000
(c) 75600
(d) None of these
Answer:
Ans. (b)
Required number of words = \( {}^{6}C_{4} \times {}^{5}C_{3} \times 7! = 15 \times 10 \times 5040 = 756000 \)
[Selection can be made in \( {}^{6}C_{4} \times {}^{5}C_{3} \) ways while the 7 letters can be arranged in 7! ways.]

Example 25. 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 [Rajasthan PET 1981]
(a) \( {}^{5}C_{3} \times {}^{22}C_{9} \)
(b) \( {}^{22}C_{9} - {}^{5}C_{3} \)
(c) \( {}^{22}C_{3} + {}^{5}C_{3} \)
(d) None of these
Answer:
Ans. (a)
The 3 reserved vacancies are filled from the 5 scheduled caste candidates in \( {}^{5}C_{3} \) ways.
Now the remaining candidates are 22 and the remaining seats are 9, so the remaining vacancies are filled in \( {}^{22}C_{9} \) ways.
Hence the total number of ways = \( {}^{5}C_{3} \times {}^{22}C_{9} \).

Example 26. 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:
Ans. (b)
Each lamp can be dealt with in 2 ways (on or off), so the number of ways = 2¹⁰ − 1 = 1023.
[1 is subtracted for the case when none of the lamps is switched on.]

Example 27. 10 different letters of English alphabet are given. Out of these letters, words of 5 letters are formed. How many words are formed when at least one letter is repeated
(a) 69760
(b) 98748
(c) 96747
(d) 97147
Answer:
Ans. (a)
Number of words of 5 letters in which letters may be repeated any number of times = 10⁵.
Number of words of 5 letters with all letters different = \( {}^{10}P_{5} = 30240 \) (a word is an arrangement, so order counts).
∴ Required number of words = 10⁵ − 30240 = 69760.

Example 28. A man has 10 friends. In how many ways can he invite one or more of them to a party
(a) 10!
(b) 2¹⁰
(c) 10! − 1
(d) 2¹⁰ − 1
Answer:
Ans. (d)
Required number of ways = 2¹⁰ − 1 (since the case that no friend is invited, i.e., \( {}^{10}C_{0} \), is excluded).

Example 29. 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:
Ans. (b)
Numbers greater than 1000 and less than or equal to 4000 will be of 4 digits and will have either 1 (except 1000) or 2 or 3 in the first place, or will be 4000.
After fixing the 1st place, the second place can be filled by any of the 5 digits. Similarly the third place can be filled in 5 ways and the 4th place in 5 ways. Thus there will be 5 × 5 × 5 = 125 ways in which 1 will be in the first place, but this includes 1000 also, hence there will be 124 numbers having 1 in the first place.
Similarly there are 125 numbers each for 2 or 3 in the first place. One number has 4 in the first place, i.e., 4000.
Hence the required numbers are 124 + 125 + 125 + 1 = 375.

Example 30. In the 13 cricket players 4 are bowlers, then in how many ways can a cricket team of 11 players be formed in which at least 2 bowlers are included
(a) 55
(b) 72
(c) 78
(d) None of these
Answer:
Ans. (c)
2 bowlers and 9 other players = \( {}^{4}C_{2} \times {}^{9}C_{9} \); 3 bowlers and 8 other players = \( {}^{4}C_{3} \times {}^{9}C_{8} \);
4 bowlers and 7 other players = \( {}^{4}C_{4} \times {}^{9}C_{7} \).
Hence required number of ways = 6 × 1 + 4 × 9 + 1 × 36 = 78.

Example 31. In how many ways can a team of 10 players out of 22 players be made if 6 particular players are always to be included and 4 particular players are always excluded
(a) \( {}^{22}C_{10} \)
(b) \( {}^{18}C_{3} \)
(c) \( {}^{12}C_{4} \)
(d) \( {}^{18}C_{4} \)
Answer:
Ans. (c)
6 particular players are always to be included and 4 are always excluded, so the remaining 4 players are to be selected out of 12.
Hence number of ways = \( {}^{12}C_{4} \).

Example 32. In how many ways can 6 persons be selected from 4 officers and 8 constables, if at least one officer is to be included [Roorkee 1985; MP PET 2001]
(a) 224
(b) 672
(c) 896
(d) None of these
Answer:
Ans. (c)
Required number of ways = \( {}^{4}C_{1} \times {}^{8}C_{5} + {}^{4}C_{2} \times {}^{8}C_{4} + {}^{4}C_{3} \times {}^{8}C_{3} + {}^{4}C_{4} \times {}^{8}C_{2} \)
= 4 × 56 + 6 × 70 + 4 × 56 + 1 × 28 = 896.

Example 33. In how many ways can 5 prizes be distributed among four students when every student can take one or more prizes [BIT Ranchi 1990; Rajasthan PET 1988, 97]
(a) 1024
(b) 625
(c) 120
(d) 60
Answer:
Ans. (a)
Each prize can be given to any one of the 4 students, i.e., in 4 ways.
By the multiplication principle, the required number of ways = 4⁵ = 1024.

Example 34. A question paper is divided into two parts A and B and each part contains 5 questions. The number of ways in which a candidate can answer 6 questions selecting at least two questions from each part is
(a) 80
(b) 100
(c) 200
(d) None of these
Answer:
Ans. (c)
2 questions from A and 4 from B = \( {}^{5}C_{2} \times {}^{5}C_{4} = 50 \); 3 questions from A and 3 from B = \( {}^{5}C_{3} \times {}^{5}C_{3} = 100 \);
4 questions from A and 2 from B = \( {}^{5}C_{4} \times {}^{5}C_{2} = 50 \).
Hence total number of ways = 50 + 100 + 50 = 200.

Example 35. The number of diagonals in an octagon will be [MP PET 1984; Pb. CET 1989, 2000]
(a) 28
(b) 20
(c) 10
(d) 16
Answer:
Ans. (b)
Number of diagonals = \( {}^{8}C_{2} - 8 = 28 - 8 = 20 \).

Example 36. The number of straight lines joining 8 points on a circle is
(a) 8
(b) 16
(c) 24
(d) 28
Answer:
Ans. (d)
No three points on a circle are collinear. Number of straight lines = \( {}^{8}C_{2} = 28 \).

Example 37. The number of triangles that can be formed by choosing the vertices from a set of 12 points, seven of which lie on the same straight line, is [Roorkee 1989, 2000; BIT Ranchi 1989; MP PET 1995; Pb. CET 1997; DCE 2000]
(a) 185
(b) 175
(c) 115
(d) 105
Answer:
Ans. (a)
Required number of ways = \( {}^{12}C_{3} - {}^{7}C_{3} = 220 - 35 = 185 \).

Example 38. 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 (ii) triangles which can be formed by joining them
(i) (a) 140   (b) 142   (c) 144   (d) 146
(ii) (a) 816   (b) 806   (c) 800   (d) 750
Answer:
Ans. (i) (c), (ii) (b)
Out of 18 points, 5 are collinear.
(i) Number of straight lines = \( {}^{18}C_{2} - {}^{5}C_{2} + 1 = 153 - 10 + 1 = 144 \)
(ii) Number of triangles = \( {}^{18}C_{3} - {}^{5}C_{3} = 816 - 10 = 806 \).

Example 39. 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 [IIT 1987; Rajasthan PET 1999]
(a) 2
(b) 3
(c) 4
(d) None of these
Answer:
Ans. (b)
Since the student is allowed to select utmost n books out of (2n + 1) books, in order to select at least one book he has the choice to select one, two, three, ……, n books.
Thus, if T is the total number of ways of selecting at least one book, then \( T = {}^{2n+1}C_{1} + {}^{2n+1}C_{2} + \ldots + {}^{2n+1}C_{n} = 63 \).
Again, the sum of all the coefficients \( {}^{2n+1}C_{0} + {}^{2n+1}C_{1} + \ldots + {}^{2n+1}C_{2n+1} = 2^{2n+1} \), and since \( {}^{2n+1}C_{r} = {}^{2n+1}C_{2n+1-r} \), this gives \( 1 + 2T + 1 = 2^{2n+1} \).
\( \Rightarrow 1 + T = 2^{2n} \Rightarrow 1 + 63 = 2^{2n} \Rightarrow 2^{6} = 2^{2n} \Rightarrow n = 3 \).