Punjab Board Class 11 Mathematics

Chapter 7: Permutations and Combinations — Class 11 Maths Notes

Permutations and combinations: counting laws applied step by step so you always pick the right formula first. This page has the theory, solved textbook examples, every exercise solution, MCQs and a formula sheet for Chapter 7.

Exercise solutions — Chapter 7

Every exercise of Permutations and Combinations solved step by step.

Exercise 7.1 Exercise 7.2 Exercise 7.3 Exercise 7.4

Mathematics 11 (Punjab Board)

Chapter 7: Permutations and Combinations — Theory

Definitions, key results and theory for Punjab Board Class 11 Mathematics, Chapter 7: Permutations and Combinations.

Factorial Notation

The product of the first \(n\) natural numbers is written as \(n!\) (read "n factorial"). Yeh counting ke tamam formulas ki bunyad hai — jab bhi objects ko arrange ya select karna ho, factorial hi kaam aata hai.

n! = n(n-1)(n-2)\cdots 3 \cdot 2 \cdot 1

Two special values to memorise: \(0! = 1\) and \(1! = 1\). Also note the useful reduction \(n! = n \cdot (n-1)!\).

\frac{n!}{(n-1)!} = n, \qquad \frac{n!}{(n-r)!} = n(n-1)\cdots(n-r+1)

Key point: In exams, always expand the larger factorial so it cancels with the smaller one — e.g. \(\frac{10!}{8!} = 10 \cdot 9 = 90\).

Fundamental Principle of Counting

If one operation can be done in \(m\) ways and a second, independent operation in \(n\) ways, the two together can be done in \(m \times n\) ways. Isay multiplication principle kehte hain — jab kaam steps mein ho aur har step ke apne options hon, to tadad multiply hoti hai.

\text{Total ways} = m_1 \times m_2 \times \cdots \times m_k

If the operations are mutually exclusive alternatives (either this or that), add the ways instead — the addition principle.

Key point: "AND" (step by step) means multiply; "OR" (this way or that way) means add.

Permutations

A permutation is an arrangement of objects in which the order matters. \(n\) different objects mein se \(r\) objects ke arrangements ki tadad \(^{n}P_{r}\) se zahir hoti hai.

^{n}P_{r} = \frac{n!}{(n-r)!} = n(n-1)(n-2)\cdots(n-r+1), \qquad 0 \le r \le n

Special cases: \(^{n}P_{n} = n!\) (arranging all objects) and \(^{n}P_{0} = 1\) (arranging nothing — exactly one way, the empty arrangement).

Key point: Permutation use karo jab order important ho — numbers banana, words banana, seats par bithana, signals dena.

Permutations of Objects Not All Different

Jab kuch objects identical hon (jaise BANANA mein A teen dafa), to unki aapas mein tabdeeli se naya arrangement nahi banta. Is repetition ko divide karke nikalte hain.

\frac{n!}{p! \, q! \, r!} \quad \text{(}p\text{ alike of one kind, }q\text{ of another, }r\text{ of a third)}

Example: MISSISSIPPI has 11 letters with I four times, S four times, P twice, so distinct arrangements \(= \frac{11!}{4!\,4!\,2!} = 34650\).

Key point: Har repeated group ke factorial se divide karo — repeated letters "overcount" ko khatam karte hain.

Circular Permutations

Gol mez ke gird bithana line mein bithane se mukhtalif hai, kyunke sirf relative positions matter karti hain. Aik shakhs ko fix kar do, baqi \((n-1)\) ko arrange karo.

(n-1)! \quad \text{(round table); } \qquad \frac{(n-1)!}{2} \quad \text{(necklace / key ring, flippable)}

A necklace or key ring can be turned over, so clockwise and anticlockwise orders are identical — divide by 2.

Key point: Round table \(\to (n-1)!\); necklace/key ring \(\to (n-1)!/2\). Sawal mein "flipped" ya "identical" ka zikr ho to 2 se divide karo.

Combinations

A combination is a selection in which the order does not matter. \(n\) objects mein se \(r\) ka intikhab \(^{n}C_{r}\) hota hai.

^{n}C_{r} = \frac{n!}{r!(n-r)!} = \frac{^{n}P_{r}}{r!}, \qquad 0 \le r \le n

Kyunki har selection ke \(r!\) arrangements hote hain, isliye \(^{n}P_{r} = ^{n}C_{r} \times r!\) — yeh relation har numerical mein kaam aata hai.

Complementary combinations: choosing \(r\) is the same as leaving \((n-r)\) behind.

^{n}C_{r} = ^{n}C_{\,n-r}, \qquad ^{n}C_{0} = ^{n}C_{n} = 1

Key point: Combination use karo jab sirf "kaun" matter kare, "kis order mein" nahi — committee banana, cards nikalna, dishes select karna.

Pascal's Identity and Useful Results

Yeh identities proofs aur short questions mein baar baar aati hain.

^{n}C_{r} + ^{n}C_{r-1} = ^{n+1}C_{r} \qquad \text{(Pascal's identity)}

r \cdot ^{n}C_{r} = (n-r+1) \cdot ^{n}C_{r-1}, \qquad ^{n}P_{r} = n \cdot ^{n-1}P_{r-1}

Diagonals of an \(n\)-sided polygon: any two vertices give a segment, minus the \(n\) sides.

\text{Diagonals} = ^{n}C_{2} - n = \frac{n(n-3)}{2}

Key point: Pehle decide karo — order matters (permutation) ya nahi (combination). Ghalat formula chunna sab se bara exam mistake hai.

Mathematics 11 (Punjab Board)

Solved Examples: Chapter 7 – Permutations and Combinations

Textbook solved examples for Punjab Board Class 11 Mathematics, Chapter 7: Permutations and Combinations, worked in full.

Example 1: Evaluate \(\frac{8!}{6!}\).

Factorial wale sawalon mein bara factorial khol kar chhote factorial se cancel kar dete hain. Yahan \(8! = 8 \cdot 7 \cdot 6!\) likhenge.

\frac{8!}{6!} = \frac{8 \cdot 7 \cdot 6!}{6!} = 8 \cdot 7 = 56

Answer: \(56\)

Example 2: Write \(8 \cdot 7 \cdot 6 \cdot 5\) in the factorial form.

Product ko factorial banane ke liye numerator aur denominator dono mein \(4!\) se multiply kar dete hain, taake upar poora \(8!\) ban jaye.

8 \cdot 7 \cdot 6 \cdot 5 = \frac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4!}{4!} = \frac{8!}{4!}

Answer: \(\frac{8!}{4!}\)

Example 3: Evaluate \(\frac{9!}{6! \cdot 3!}\).

Yeh \(^{9}C_{3}\) ki hi shakal hai. \(9!\) ko \(9 \cdot 8 \cdot 7 \cdot 6!\) tak kholen, \(6!\) cancel hoga aur \(3! = 6\) se divide karenge.

\frac{9!}{6! \cdot 3!} = \frac{9 \cdot 8 \cdot 7 \cdot 6!}{6! \cdot 6} = \frac{504}{6} = 84

Answer: \(84\)

Example 4: How many different 4-digit numbers can be formed out of the digits 1, 2, 3, 4, 5, 6, when no digit is repeated?

4-digit number banana matlab 6 digits mein se 4 ko order mein chunna — order matters, isliye permutation \(^{6}P_{4}\) lagegi.

^{6}P_{4} = \frac{6!}{(6-4)!} = \frac{6!}{2!} = 6 \cdot 5 \cdot 4 \cdot 3 = 360

Answer: \(360\)

Example 5: In how many ways can a set of 4 different mathematics books, 3 different physics books and 2 different chemistry books be placed on a shelf with a space for 9 books, if: (a) all the books are kept without any restriction. (b) all the books of the same subject are kept together. (c) only the mathematics books are kept together.

Jin books ko sath rakhna ho unhein aik "block" bana lo, phir blocks ko arrange karo aur block ke andar ki tartib se multiply karo.

(a) No restriction:

\[ 9! = 362880 \]

(b) Same subject together:

3! \times 4! \times 3! \times 2! = 6 \times 24 \times 6 \times 2 = 1728

(c) Only mathematics books together:

6! \times 4! = 720 \times 24 = 17280

Answer: (a) \(362880\), (b) \(1728\), (c) \(17280\)

Example 6: In how many ways can 5 people be seated on a bench if: (a) there are no restrictions (b) two particular people sit next to each other (c) two particular people cannot sit next to each other.

Bench par bithana linear arrangement hai. "Sath bithana" ho to dono ko aik unit samjho; "sath na bithana" ho to total mein se sath wale cases minus kar do.

(a) No restrictions:

\[ 5! = 120 \]

(b) Two particular people together:

4! \times 2! = 24 \times 2 = 48

(c) Two particular people not together:

\[ 120 - 48 = 72 \]

Answer: (a) \(120\), (b) \(48\), (c) \(72\)

Example 7: In how many ways can the letters of the word MISSISSIPPI be arranged when all the letters are to be used?

Letters repeat ho rahe hain: 11 letters mein I 4 dafa, S 4 dafa, P 2 dafa. Isliye \(\frac{11!}{4!\,4!\,2!}\) ka formula lagega.

\frac{11!}{4! \cdot 4! \cdot 2!} = \frac{39916800}{24 \cdot 24 \cdot 2} = \frac{39916800}{1152} = 34650

Answer: \(34650\)

Example 8: In how many ways can 4 persons be seated at a round table, while: (i) clockwise and anticlockwise orders are different (ii) clockwise and anticlockwise orders are identical.

Gol mez par \((n-1)!\) arrangements hote hain. Agar clockwise aur anticlockwise aik jaise hon (jaise flip ho sake) to 2 se divide kar do.

(i) Orders different:

\[ (4-1)! = 3! = 6 \]

(ii) Orders identical:

\frac{(4-1)!}{2} = \frac{6}{2} = 3

Answer: (i) \(6\), (ii) \(3\)

Example 9: Zain has 8 different fruits. He wants to select 5 fruits out of 8 fruits to make a fruit chaat. How many combinations of fruits can he select?

Fruits "select" karne hain, arrange nahi — order irrelevant hai, isliye combination \(^{8}C_{5}\) lagegi.

^{8}C_{5} = \frac{8!}{5!\,3!} = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} = 56

Answer: \(56\)

Example 10: In a school, a class consists of 12 girls and 8 boys. The teacher wants to select 5 students for an activity. In how many ways can the students be selected including: (i) 2 girls (ii) 5 boys (iii) 2 boys.

Girls aur boys ke selections alag alag honge, phir multiplication principle se multiply karenge.

(i) 2 girls and 3 boys:

^{12}C_{2} \times ^{8}C_{3} = 66 \times 56 = 3696

(ii) 5 boys:

^{8}C_{5} = 56

(iii) 2 boys and 3 girls:

^{8}C_{2} \times ^{12}C_{3} = 28 \times 220 = 6160

Answer: (i) \(3696\), (ii) \(56\), (iii) \(6160\)

Example 11: Find the number of diagonals of a 6-sided figure.

Har 2 vertices aik segment banate hain (\(^{6}C_{2}\)), jin mein se 6 sides hain — baqi diagonals hongi.

^{6}C_{2} - 6 = 15 - 6 = 9

Answer: \(9\)

Example 12: Zain wants to generate a password for his laptop to secure the data. He can take only 6 characters to generate a password. Each character can either be an upper case letter (A–Z) or a digit (0–9). Find how many passwords can be generated: (i) if repetition of characters is not allowed (ii) if repetition of characters is allowed.

26 letters + 10 digits = 36 characters. Password mein order matters hai. Repetition na ho to permutation; repetition ho to har position par 36 options.

(i) Repetition not allowed:

^{36}P_{6} = 36 \cdot 35 \cdot 34 \cdot 33 \cdot 32 \cdot 31 = 1402410240

(ii) Repetition allowed:

36^{6} = 2176782336

Answer: (i) \(1402410240\), (ii) \(2176782336\)

Example 13: A box contains 15 cards numbered 1–15. Danish is to select 5 cards. If all the selected cards are the first five multiples of 2 then Danish will win the game. Find Danish's chance of winning the game, when (i) order is important (ii) order is not important.

Jeetne wala sirf aik set hai \(\{2,4,6,8,10\}\). Probability = favourable ÷ total. Order important ho to permutation, warna combination.

(i) Order important:

\frac{1}{^{15}P_{5}} = \frac{1}{360360}

(ii) Order not important:

\frac{1}{^{15}C_{5}} = \frac{1}{3003}

Answer: (i) \(\frac{1}{360360}\), (ii) \(\frac{1}{3003}\)

Example 14: On Independence Day, a DJ has a list of ten different national songs. He wants to select any five national songs for the day. Find how many ways he can select and play the songs: (i) if the order of playing the songs matters (ii) if the order of playing the songs does not matter.

"Select and play" — agar bajane ki tartib matter kare to permutation, sirf chunao ho to combination.

(i) Order matters:

^{10}P_{5} = 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 = 30240

(ii) Order does not matter:

^{10}C_{5} = \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 252

Answer: (i) \(30240\), (ii) \(252\)

Mathematics 11 (Punjab Board)

MCQs: Chapter 7 – Permutations and Combinations

If a student has 4 different shirts and 3 different pants, in how many ways can he choose an outfit?
1. 7
2. 10
3. 12
4. 24
Multiplication principle: \(4 \times 3 = 12\). Har shirt ke sath 3 pants pehni ja sakti hain.
Evaluate: \(\frac{8!}{6!}\)
1. 2
2. 48
3. 56
4. 72
\(\frac{8 \cdot 7 \cdot 6!}{6!} = 8 \cdot 7 = 56\).
If \(^{n}P_{2} = 30\), find the value of \(n\).
1. 5
2. 6
3. 7
4. 8
\(n(n-1) = 30 = 6 \cdot 5\), isliye \(n = 6\).
How many distinct ways can the letters of the word "PYTHON" be arranged?
1. 120
2. 720
3. 5040
4. 360
6 different letters: \(6! = 720\).
The number of distinct permutations of the letters of the word "MISSISSIPPI" is:
1. \(\frac{11!}{4! \cdot 4!}\)
2. \(\frac{11!}{4! \cdot 4! \cdot 2!}\)
3. \(\frac{11!}{4! \cdot 2!}\)
4. \(11!\)
11 letters: I 4 dafa, S 4 dafa, P 2 dafa — \(\frac{11!}{4!\,4!\,2!}\).
In how many ways can 5 people be seated around a round table?
1. 120
2. 24
3. 60
4. 5
Circular permutation: \((5-1)! = 24\).
If clockwise and anticlockwise arrangements are considered identical, how many ways can 4 beads of different colours be arranged in a necklace?
1. 6
2. 12
3. 3
4. 24
Necklace flip ho sakta hai: \(\frac{(4-1)!}{2} = 3\).
Evaluate: \(^{5}C_{2}\)
1. 20
2. 5
3. 10
4. 15
\(\frac{5 \cdot 4}{2 \cdot 1} = 10\).
If \(^{n}C_{3} = ^{n}C_{10}\), find the value of \(n\).
1. 7
2. 30
3. 13
4. 10
\(^{n}C_{r} = ^{n}C_{s} \Rightarrow n = r + s = 3 + 10 = 13\).
What is the number of diagonals of a hexagon?
1. 6
2. 15
3. 9
4. 12
\(\frac{6(6-3)}{2} = 9\).
In how many ways can a committee of 3 be selected from 10 people?
1. 720
2. 120
3. 30
4. 10
Order nahi matter karta: \(^{10}C_{3} = 120\).
Zain wants to generate a 6-character password using 26 letters and 10 digits. If repetition is not allowed, how many passwords are possible?
1. \(36^{6}\)
2. \(^{36}P_{6}\)
3. \(^{36}C_{6}\)
4. \(6!\)
Password mein order matters aur repetition nahi: \(^{36}P_{6}\).
A box contains 15 cards. If 5 cards are selected, what is the probability of winning if order does not matter?
1. \(\frac{1}{360360}\)
2. \(\frac{1}{3003}\)
3. \(\frac{1}{15!}\)
4. \(\frac{1}{5!}\)
Total selections \(^{15}C_{5} = 3003\), favourable 1: probability \(\frac{1}{3003}\).
In how many ways can 4 different mathematics books and 3 different physics books be placed on a shelf such that all books of the same subject are together?
1. \(4! \times 3!\)
2. \(2! \times 4! \times 3!\)
3. \(7!\)
4. \(4! + 3!\)
2 blocks (\(2!\)) aur har block ke andar ki tartib: \(2! \times 4! \times 3!\).
What is the value of \(^{n}C_{n}\)?
1. \(n\)
2. \(n!\)
3. 1
4. 0
Sab ko select karne ka aik hi tareeqa: \(^{n}C_{n} = 1\).
If \(^{n}P_{0} = 1\), what does this represent?
1. Selecting all objects
2. Arranging zero objects
3. Arranging all objects
4. Selecting zero objects
Zero objects ko arrange karne ka aik tareeqa — khaali arrangement.
A polygon has 104 diagonals. How many sides does it have?
1. 12
2. 14
3. 16
4. 18
\(\frac{n(n-3)}{2} = 104 \Rightarrow n^{2} - 3n - 208 = 0 \Rightarrow n = 16\).
In how many ways can 2 girls and 3 boys be selected from 12 girls and 8 boys?
1. \(^{20}C_{5}\)
2. \(^{12}P_{2} \times ^{8}P_{3}\)
3. \(^{12}C_{2} \times ^{8}C_{3}\)
4. \(^{12}C_{3} \times ^{8}C_{2}\)
Dono selections alag: \(^{12}C_{2} \times ^{8}C_{3}\).
If 5 flags of different colours are available, how many signals can be made using any number of flags at a time?
1. \(2^{5} - 1\)
2. \(^{5}P_{1} + ^{5}P_{2} + ^{5}P_{3} + ^{5}P_{4} + ^{5}P_{5}\)
3. \(5!\)
4. \(2^{5}\)
Signal mein order matters: har mumkin tadad ki permutations ka sum \(= 325\).
The number of ways to arrange the letters of the word "BANANA" such that B is always the first letter is:
1. \(\frac{6!}{3! \cdot 2!}\)
2. \(\frac{5!}{3! \cdot 2!}\)
3. \(\frac{5!}{3!}\)
4. \(5!\)
B fix — baqi 5 letters (A 3 dafa, N 2 dafa): \(\frac{5!}{3!\,2!} = 10\).

Mathematics 11 (Punjab Board)

Chapter 7: Permutations and Combinations — Formula Sheet

Factorials

\[ n! = n(n-1)(n-2)\cdots 3 \cdot 2 \cdot 1 \]

\[ 0! = 1, \qquad 1! = 1 \]

\[ \frac{n!}{(n-1)!} = n, \qquad \frac{(n+1)!}{n!} = n+1 \]

Fundamental Principle of Counting

\[ \text{Multiplication principle: } m_1 \times m_2 \times \cdots \times m_k \]

\[ \text{Addition principle: } m + n \text{ (mutually exclusive alternatives)} \]

Permutations (order matters)

\[ ^{n}P_{r} = \frac{n!}{(n-r)!} \]

\[ ^{n}P_{n} = n!, \qquad ^{n}P_{0} = 1 \]

\[ ^{n}P_{r} = n \cdot ^{n-1}P_{r-1} \]

Permutations of Objects Not All Different

\[ \frac{n!}{p!\,q!\,r!} \]

Circular Permutations

\[ (n-1)! \quad \text{(round table)} \]

\[ \frac{(n-1)!}{2} \quad \text{(necklace, key ring — flippable)} \]

Combinations (order does not matter)

\[ ^{n}C_{r} = \frac{n!}{r!(n-r)!} = \frac{^{n}P_{r}}{r!} \]

\[ ^{n}C_{r} = ^{n}C_{\,n-r}, \qquad ^{n}C_{0} = ^{n}C_{n} = 1 \]

\[ ^{n}C_{r} + ^{n}C_{r-1} = ^{n+1}C_{r} \quad \text{(Pascal's identity)} \]

\[ r \cdot ^{n}C_{r} = (n-r+1) \cdot ^{n}C_{r-1} \]

Geometry Applications

\[ \text{Diagonals of an n-gon} = \frac{n(n-3)}{2} \]

\[ \text{Triangles from n vertices} = ^{n}C_{3} \]

Chapter 7 FAQs

What is covered in Chapter 7 (Permutations and Combinations)?

This chapter covers Factorials, permutations, combinations, multiplication and addition laws. The page includes theory and definitions, solved textbook examples, complete exercise solutions, MCQs and a formula sheet.

How many marks does Chapter 7 carry in the Punjab Board paper?

Chapter 7 (Permutations and Combinations) carries approximately 13 marks in the Punjab Board Class 11 Mathematics paper.

Are these notes for the new PCTB syllabus?

Yes. These notes follow the PCTB 2025 new book for Class 11 Mathematics, used by BISE Lahore, Gujranwala, Rawalpindi, Faisalabad, Multan, Sahiwal, Sargodha, Bahawalpur and D.G. Khan.

Where are the exercise solutions for Chapter 7?

Use the exercise index on this page — every exercise of Chapter 7 has its own page with complete step-by-step solutions.