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NCERT Solutions for Class 11 Maths Chapter 6 Permutations And Combinations

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Master Permutations And Combinations Class 11 Questions And Answers for Higher Marks

NCERT Solutions for Chapter 6 of Class 11 Maths covers Permutations and Combinations, crucial concepts for counting arrangements and selections. This chapter helps students understand how to count objects in a specific order (permutations) and how to choose objects without considering the order (combinations). These principles are essential for solving various mathematical problems and have practical applications in probability, statistics, and real-life scenarios. By studying this chapter, students will develop strong analytical skills and the ability to solve complex counting problems efficiently. Vedantu’s Class 11 Maths NCERT Solutions provide clear, step-by-step explanations to help you master these concepts.

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Master Permutations And Combinations Class 11 Questions And Answers for Higher Marks

Exercise 6.1: This exercise consists of 8 questions and introduces students to the concept of permutations. The questions are based on various concepts such as the meaning of permutations, the formula for finding the number of permutations of n objects taken r at a time, and solving problems related to permutations. Students will learn how to calculate the number of permutations and solve problems related to permutations.

Exercise 6.2: This exercise consists of 12 questions and is focused on solving problems related to permutations and combinations. The questions are based on various concepts such as the difference between permutations and combinations, the formula for finding the number of combinations of n objects taken r at a time, and solving problems related to combinations. Students will learn how to calculate the number of combinations and solve problems related to permutations and combinations.

Exercise 6.3: This exercise consists of 13 questions and is focused on solving problems related to the application of permutations and combinations. The questions are based on various concepts such as the principle of multiplication and addition, solving problems related to the selection of objects, and solving problems related to the distribution of objects. Students will learn how to apply the concepts of permutations and combinations to solve problems related to various scenarios.

Exercise 6.4: This exercise consists of 10 questions and is focused on solving problems related to the application of permutations and combinations in probability. The questions are based on various concepts such as the meaning of probability, solving problems related to the probability of an event, and solving problems related to independent and dependent events. Students will learn how to calculate probabilities using the concepts of permutations and combinations.

Miscellaneous Exercise: This exercise consists of 11 questions and covers a variety of topics related to permutations and combinations. The questions are based on various concepts such as solving problems related to the number of ways to select objects, finding the value of n, solving problems related to the arrangement of objects, and solving word problems related to permutations and combinations. This exercise will help students to revise and reinforce the concepts learned in the previous exercises.


Access NCERT Solutions for Class 11 Maths Chapter 6 – Permutations and Combinations

Exercise 6.1

1. How many 3 -digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that
(i) Repetition of the digits is allowed?
(ii) Repetition of the digits is not allowed?

Ans: (i) There are as many ways of filling 3 empty places in succession by the given 5 digits. In the given case, repetition of digits is allowed. Thus, the units place can be filled in by any of the all 5 digits. Similarly, tens and hundreds digits can be filled by any of the all 5 digits. Thus, the number of ways in which 3-digit numbers can be formed from the given digits is $5 \times 5 \times 5\, = \,125.$
(ii) In the given case, repetition of digits is not allowed. Thus, if units place is filled at first, then it can be filled by any of the given 5 digits. Thus, the number of ways is 5 . Similarly, the tens place can be filled by remaining 4 digits and the hundreds place can be filled by remaining three digits. Therefore, by multiplication principle, the number of ways = $5 \times 4 \times 3\, = \,60.$


2. How many 3-digit even numbers cab be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?

Ans: There are ways of filling 3 empty places in succession by the given 6 digits.


In the given case, the units place can be filled in by any of the all 6 digits. Similarly, tens and hundreds digits can be filled by any of the all 6 digits, as repetition is allowed.


Thus, by multiplication principle, the required 3-digit even numbers is

$3 \times 6 \times 6\, = \,108$


3. How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?

Ans: There are ways of filling 4 empty places in succession by the given 10 letters of the English alphabet, given that repetition is not allowed. The first place can be filled in 10 different ways any of the given 10 letters and second place can be filled in 9 different ways. Similarly third place can be filled in 8 different ways and the fourth in 7 different ways.


Thus, by multiplication principle, the number of ways in which 4 empty places can be filled is $10 \times 9 \times 8 \times 7\, = \,5040$


Therefore, 5040 4-letter codes can be formed with the help of first 10 letters of English alphabet without repetition.


4. How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?

Ans:  Given: 5-digit telephone number starts with 67.


Therefore, there are ways of filling the 3 empty spaces 6, 7, __, __, __ by digits 0-9 without repetition.


Thus, units place can be filled in 8 different ways, tens place can be filled in 7 different ways and hundreds place can be filled in 6 different ways.


Thus, by multiplication principle, number of ways in which 5-digit telephone number can be constructed is $8 \times 7 \times 6\, = \,336$


5. A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?

Ans: When a coin is tossed once, there are 2 outcomes (head and tail) i.e., the number of ways of showing different face = 2


Thus, by multiplication principle, number of possible outcomes = $2 \times 2 \times 2\, = \,8$


6. Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?

Ans: 

1 signal = 2 flags.


There are ways of filling 2 empty places in succession by the given 5 flags.


The upper empty place can be filled in 5 different ways and the lower empty place can be filled in 4 different ways.


Thus, by the multiplication principle, the ways in which different flags can be generated =$5 \times 4\, = \,20$


Exercise 6.2

1. Evaluate (i) 8!

Ans: ${\text{(i)}}{\mkern 1mu} \,{\text{8!}}{\mkern 1mu} {\text{  =  }}{\mkern 1mu} 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8{\mkern 1mu} \, = \,40320$

(ii) 4! – 3!

Ans: ${\text{4!   =   1}} \times {\text{2}} \times {\text{3}} \times {\text{4}}{\mkern 1mu} {\text{  =  }}{\mkern 1mu} {\text{24}}  $

$  3!{\mkern 1mu} \, = {\mkern 1mu} {\text{1}} \times {\text{2}} \times {\text{3}}{\mkern 1mu} {\text{  =  }}{\mkern 1mu} {\text{6}}  $

 $ \therefore {\text{4!}}{\mkern 1mu} {\text{  -  }}{\mkern 1mu} {\text{3!}}{\mkern 1mu} {\text{ }}  $

  ${\text{ =  }}{\mkern 1mu} {\text{24}}{\mkern 1mu} {\text{  -  }}{\mkern 1mu} {\text{6}}{\mkern 1mu} {\text{ }}  $

 $ {\text{ =  }}{\mkern 1mu} {\text{18}}  $

 

2. Is 4! + 3! = 7!?

Ans: $ 3!{\mkern 1mu}  = {\mkern 1mu} {\text{1}} \times {\text{2}} \times {\text{3}}{\mkern 1mu} {\text{  =  }}{\mkern 1mu} {\text{6}}  $

$  {\text{4! = 1}} \times {\text{2}} \times {\text{3}} \times {\text{4}}{\mkern 1mu} {\text{  =  }}{\mkern 1mu} {\text{24}}  $

 $ \therefore 3!{\mkern 1mu}  + {\mkern 1mu} 4! = {\mkern 1mu} 6 + 24{\mkern 1mu}  = {\mkern 1mu} 30  $

$  7!{\mkern 1mu}  = {\mkern 1mu} {\text{1}} \times {\text{2}} \times {\text{3}} \times {\text{4}} \times 5 \times 6 \times 7{\mkern 1mu}  = {\mkern 1mu} 5040  $

$  \therefore 3!{\mkern 1mu}  + {\mkern 1mu} {\text{4!}}{\mkern 1mu}  \ne {\mkern 1mu} 7!  $


3. Compute $\dfrac{{8!}}{{6!{\mkern 1mu}  \times {\mkern 1mu} 2!}}$

Ans:  \[\dfrac{{8!}}{{6!{\mkern 1mu}  \times {\mkern 1mu} 2!}} = \dfrac{{8 \times 7 \times 6!}}{{6! \times 2 \times 1}} = \dfrac{{8 \times 7}}{2} = 28\]

 

4. If $\dfrac{1}{{6!}}{\mkern 1mu}  + {\mkern 1mu} \dfrac{1}{{7!}}{\mkern 1mu}  = {\mkern 1mu} \dfrac{{\text{x}}}{{8!}}$, find x.

Ans: 

$\dfrac{1}{{6!}}{\mkern 1mu}  + {\mkern 1mu} \dfrac{1}{{7!}}{\mkern 1mu}  = {\mkern 1mu} \dfrac{x}{{8!}}  $

  $ \Rightarrow \dfrac{1}{{6!}} + \dfrac{1}{{7! \times 6!}}{\mkern 1mu}  = {\mkern 1mu} \dfrac{x}{{8 \times 7 \times 6!}}  $

  $ \Rightarrow \dfrac{1}{{6!}}\left( {1{\mkern 1mu}  + {\mkern 1mu} \dfrac{1}{7}} \right){\mkern 1mu}  = {\mkern 1mu} \dfrac{x}{{8 \times 7 \times 6!}}  $

$   \Rightarrow 1{\mkern 1mu}  + {\mkern 1mu} \dfrac{1}{7}{\mkern 1mu}  = {\mkern 1mu} \dfrac{x}{{8 \times 7}}  $

 $  \Rightarrow \dfrac{8}{7}{\mkern 1mu}  = {\mkern 1mu} \dfrac{x}{{8 \times 7}}  $

  $x{\mkern 1mu} {\text{ }} = {\text{ }}{\mkern 1mu} \dfrac{{8 \times 8 \times 7}}{7}  $

 $ x{\mkern 1mu} {\text{ }} = {\text{ }}{\mkern 1mu} 64  $


5.Evaluate $\dfrac{{{\text{n!}}}}{{{\text{(n}}{\mkern 1mu} {\text{  -  }}{\mkern 1mu} {\text{r)!}}}},$when 

(i) n = 6, r = 2 (ii) n = 9, r = 5

Ans: $ {\text{(i)}}{\mkern 1mu} {\text{When}}{\mkern 1mu} {\text{n  =  6,}}{\mkern 1mu} {\text{r  =  2:}}  $

$  \dfrac{{{\text{n!}}}}{{{\text{(n}}{\mkern 1mu} {\text{  -  }}{\mkern 1mu} {\text{r)!}}}}{\mkern 1mu} {\text{  =  }}{\mkern 1mu} \dfrac{{{\text{6!}}}}{{{\text{(6}}{\mkern 1mu} {\text{  -  }}{\mkern 1mu} {\text{2)!}}}}{\text{ }}  $

$  {\text{ =  }}\dfrac{{{\text{6!}}}}{{{\text{4!}}}}{\text{  =  }}\dfrac{{{\text{6}} \times {\text{5}} \times {\text{4}}!}}{{{\text{4!}}}}{\text{  =  30}}  $

$  {\text{(ii)}}{\mkern 1mu} {\text{When}}{\mkern 1mu} {\text{n  =  9,}}{\mkern 1mu} {\text{r  =  5:}}  $

$  \dfrac{{{\text{n!}}}}{{{\text{(n}}{\mkern 1mu} {\text{  -  }}{\mkern 1mu} {\text{r)!}}}}{\mkern 1mu} {\text{  =  }}{\mkern 1mu} \dfrac{{{\text{9!}}}}{{{\text{(9}}{\mkern 1mu} {\text{  -  }}{\mkern 1mu} {\text{5)!}}}}{\mkern 1mu}   $

$  {\text{  =  }}{\mkern 1mu} \dfrac{{{\text{9!}}}}{{{\text{4!}}}}{\mkern 1mu} {\text{  =  }}{\mkern 1mu} \dfrac{{{\text{9}} \times {\text{8}} \times {\text{7}} \times {\text{6}} \times {\text{5}} \times {\text{4}}!}}{{{\text{4!}}}}  $

  ${\text{ =  }}{\mkern 1mu} {\text{9}} \times {\text{8}} \times {\text{7}} \times {\text{6}} \times {\text{5}}{\mkern 1mu} {\text{  =  }}{\mkern 1mu} {\text{15120}}  $

 

Exercise 6.3

1. How many 3-digit numbers can be formed by using the digits 1 to 9 if no digits is repeated?

Ans: Here, what matters is the order of digits.


Thus, there are permutations of 9 different digits taken 3 at a time.


Therefore, the number of 3-digit numbers = 

\[^{\text{9}}{{\text{P}}_{\text{3}}}{\mkern 1mu} {\text{  =  }}{\mkern 1mu} \dfrac{{{\text{9!}}}}{{{\text{(9}}{\mkern 1mu} {\text{  -  }}{\mkern 1mu} {\text{3)!}}}}{\text{ =  }}\dfrac{{{\text{9!}}}}{{{\text{6!}}}}{\text{ =  }}\dfrac{{9 \times 8 \times 7 \times 6!}}{{6!}} = 9 \times 8 \times 7 = 504\]

 

2. How many 4-digit numbers are there with no digit repeated?

Ans: The thousands place of the 4-digit number can be filled with any digit from 1  to 9 and 0 is not included. Thus, number of ways in which thousands place is filled is 9.


Also, the hundreds, tens and units place can be filled with any digit from 0 to 9. Since, te digits cannot be repeated and thousands place is already occupied by the digit. The hundreds, tens and units place can be filled by remaining 9 digits.


Thus, there are permutations of 9 different digits taken 3 at a time.


Number of 3-digit numbers =  

$^{\text{9}}{{\text{P}}_{{\text{3}}{\mkern 1mu} }}{\text{  =  }}{\mkern 1mu} \dfrac{{{\text{9!}}}}{{{\text{(9}}{\mkern 1mu} {\text{  -  }}{\mkern 1mu} {\text{3)!}}}}{\text{  =  }}\dfrac{{{\text{9!}}}}{{{\text{6!}}}} = {\text{ }}\dfrac{{9 \times 8 \times 7 \times 6!}}{{6!}} = 9 \times 8 \times 7 = 504\;$

By, multiplication principle, number of 4-digit numbers is $9 \times 504{\mkern 1mu}  = {\mkern 1mu} 4536$

 

3. How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?

Ans: Using the numbers, 1, 2, 3, 4, 6 and 7, 3-digit numbers can be formed.


The units place can be filled by any of the digits 2, 4 or 6. Hence, there are 3 ways.


Since, it is given that, digits cannot be repeated, units place is already occupied, the hundreds and tens place can be occupied by remaining 5 digits.


Thus, number of ways of filling hundreds and tens place = 

$^5{{\text{P}}_{2{\mkern 1mu} }}{\text{ }} = {\text{ }}{\mkern 1mu} \dfrac{{5!}}{{(5{\mkern 1mu} {\text{ }} - {\text{ }}{\mkern 1mu} 2)!}}{\text{ }} = {\text{ }}\dfrac{{5!}}{{3!}} = {\text{ }}\dfrac{{5 \times 4 \times 3!}}{{3!}}{\text{ }} = {\text{ }}20$

Therefore, by multiplication principle, number of 3-digit numbers = $3 \times 20{\mkern 1mu}  = {\mkern 1mu} 60$

 

4. Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?

Ans: From the digits 1,2 ,3 ,4 and 5, 4-digit numbers can be formed.


There are permutations of 5 different things taken 4 at a time.


Thus, number of 4 digit numbers =

$^5{{\text{P}}_4}{\mkern 1mu} {\text{  =  }}{\mkern 1mu} \dfrac{{{\text{5!}}}}{{{\text{(5}}{\mkern 1mu} {\text{  -  }}{\mkern 1mu} {\text{4)!}}}}{\mkern 1mu} {\text{ }} = {\text{ }}{\mkern 1mu} \dfrac{{5!}}{{1!}} = 1 \times 2 \times 3 \times 4 \times 5{\mkern 1mu} {\text{ }} = {\text{ }}{\mkern 1mu} 120$

Out of 1, 2, 3, 4, 5, we know that even numbers end either by 2 or 4.


Thus, ways in which units place can be filled is 2.


Since, repetition is not allowed, units place is already occupied by a digit and remaining vacant places can be filled by remaining 4 digits.


Thus, number of ways in which remaining places can be filled = 

$^{\text{4}}{{\text{P}}_{\text{3}}}{\mkern 1mu} {\text{ }} = {\text{ }}{\mkern 1mu} \dfrac{{4!}}{{(4{\mkern 1mu} {\text{ }} - {\text{ }}{\mkern 1mu} 3)!}}{\mkern 1mu} {\text{ }} = {\text{ }}{\mkern 1mu} \dfrac{{4!}}{{1!}} = 4 \times 3 \times 2 \times 1{\mkern 1mu} {\text{ }} = {\text{ }}{\mkern 1mu} 24$

Therefore, by multiplication principle, number of even numbers = $24 \times 2{\mkern 1mu}  = {\mkern 1mu} 48$

 

5. From a committee of 8 persons, how many ways can we choose a chairman and a vice chairman assuming one person cannot hold more than one position.

Ans: In a committee of 8 persons, a chairman and vice chairman are selected in such away that one person can hold only one position.


Thus, ways of choosing a chairman and vice chairman is permutation of 8 objects taken 2 at a time


Therefore, number of ways 

\[^8{{\text{P}}_2}{\text{ =  }}{\mkern 1mu} \dfrac{{{\text{8!}}}}{{{\text{(8}}{\mkern 1mu} {\text{  -  }}{\mkern 1mu} {\text{2)!}}}}{\mkern 1mu} {\text{  =  }}{\mkern 1mu} \dfrac{{{\text{8!}}}}{{{\text{6!}}}}{\mkern 1mu} {\text{  =  }}{\mkern 1mu} \dfrac{{{\text{8}} \times {\text{7}} \times {\text{6}}!}}{{{\text{6!}}}}{\mkern 1mu} {\text{  =  }}{\mkern 1mu} {\text{56}}\]

 

6. Find n if $^{{\text{n  -  1}}}{{\text{P}}_{\text{3}}}{\mkern 1mu} {\text{:}}{{\mkern 1mu} ^{\text{n}}}{{\text{P}}_{\text{4}}}{\mkern 1mu} {\text{  =  }}{\mkern 1mu} 1:9$

Ans: $^{{\text{n  -  1}}}{{\text{P}}_{\text{3}}}{\mkern 1mu} {\text{:}}{{\mkern 1mu} ^{\text{n}}}{{\text{P}}_{\text{4}}}{\mkern 1mu} {\text{  =  }}{\mkern 1mu} 1:9$

$ \dfrac{{^{n - 1}{{\text{P}}_3}}}{{^n{{\text{P}}_4}}}{\text{ =  }}{\mkern 1mu} \dfrac{{\text{1}}}{{\text{9}}}$

 $  \Rightarrow \dfrac{{\left[ {\dfrac{{{\text{(n}}{\mkern 1mu} {\text{  -  }}{\mkern 1mu} {\text{1)!}}}}{{{\text{(n}}{\mkern 1mu} {\text{  -  }}{\mkern 1mu} {\text{1}}{\mkern 1mu} {\text{  -  }}{\mkern 1mu} {\text{3)!}}}}} \right]}}{{\left[ {\dfrac{{{\text{(n}}{\mkern 1mu} {\text{  -  }}{\mkern 1mu} {\text{1)!}}}}{{{\text{(n}}{\mkern 1mu} {\text{  -  }}{\mkern 1mu} {\text{4)!}}}}} \right]}}{\mkern 1mu} {\text{  =  }}{\mkern 1mu} \dfrac{{\text{1}}}{{\text{9}}}$

  $ \Rightarrow \dfrac{{{\text{(n}}{\mkern 1mu} {\text{  -  }}{\mkern 1mu} {\text{1)!}}}}{{{\text{(n}}{\mkern 1mu} {\text{  -  }}{\mkern 1mu} {\text{4)!}}}} \times \dfrac{{{\text{(n}}{\mkern 1mu} {\text{  -  }}{\mkern 1mu} {\text{4)!}}}}{{{\text{n!}}}}{\mkern 1mu} {\text{  =  }}{\mkern 1mu} \dfrac{{\text{1}}}{{\text{9}}}$

$ \dfrac{{{\text{(n}}{\mkern 1mu} {\text{  -  }}{\mkern 1mu} {\text{1)!}}}}{{{\text{n}} \times ({\text{n}}{\mkern 1mu} {\text{  -  1)!}}}}{\mkern 1mu} {\text{  =  }}{\mkern 1mu} \dfrac{{\text{1}}}{{\text{9}}}$

 $  \Rightarrow \dfrac{{\text{1}}}{{\text{n}}}{\mkern 1mu} {\text{  =  }}{\mkern 1mu} \dfrac{{\text{1}}}{{\text{9}}}\therefore {\text{n}}{\mkern 1mu} {\text{  =  }}{\mkern 1mu} {\text{9}}  $

 

7. Find r if 

${\text{ (i)}}{{\text{ }}^5}{P_r} = {2^6}{P_{r - 1}}$

Ans: $ { \Rightarrow \dfrac{{5!}}{{(5 - r)!}} = 2 \times \dfrac{{6!}}{{(6 - r + 1)!}}} $

 $ { \Rightarrow \dfrac{{5!}}{{(5 - r)!}} = \dfrac{{2 \times 6!}}{{(7 - r)!}}} $

  ${ \Rightarrow \dfrac{{5!}}{{(5 - r)!}} = \dfrac{{6 \times 5!}}{{(7 - r)(6 - r)(5 - r)!}}}$

  ${ \Rightarrow 1 = \dfrac{{2 \times 6}}{{(7 - r)(6 - r)}}}$

$ \Rightarrow (7 - r)(6 - r) = 12$

$ \Rightarrow 42 - 6r - 7r + r = 12$

$ \Rightarrow {r^2} - 13r + 30 = 0$

$ \Rightarrow {r^2} - 3r + 10r + 30 = 0$

$ \Rightarrow r(r - 3) - 10(r - 3) = 0$

$ \Rightarrow (r - 3)(r - 10) = 0$

$ \Rightarrow r = 3 = 0$ or $(r - 10) = 0$

$ \Rightarrow r = 3$ or $r = 10$

It is known that, $^n{P_r} = \dfrac{{n!}}{{(n - r)!}}$, where $0 \leqslant r \leqslant n$

${\therefore 0 \leqslant r \leqslant 5}  $

$ {\therefore r = 3}  $

$  {\text{ (ii)}}{{\text{ }}^5}{P_r}{ = ^6}{P_{r - 1}}  $

$  {\text{Ans:}}  $

$  { \Rightarrow \dfrac{{5!}}{{(5 - r)!}} = \dfrac{{6!}}{{(6 - r + 1)!}}}  $

$  { \Rightarrow \dfrac{{5!}}{{(5 - r)!}} = \dfrac{{6 \times 5!}}{{(7 - r)!}}}  $

$  { \Rightarrow \dfrac{{5!}}{{(5 - r)!}} = \dfrac{6}{{(7 - r)(6 - r)(5 - r)!}}}$

$  { \Rightarrow 1 = \dfrac{6}{{(7 - r)(6 - r)}}}  $

  ${ \Rightarrow (7 - r)(6 - r) = 6}  $

  ${ \Rightarrow 42 - 7r - 6r + {r^2} - 6 = 0}  $

$  { \Rightarrow {r^2} - 13r + 36 = 0}  $

 $ { \Rightarrow {r^2} - 4r - 9r(r - 4) = 0}  $

 $ { \Rightarrow (r - 4)(r - 9) = 0}  $

 $ { \Rightarrow (r - 4)(r - 9) = 0}  $

$  { \Rightarrow (r - 4) = 0{\text{ or }}(r - 9) = 0}  $

 $ { \Rightarrow r = 4{\text{ or }}r = 9}$

It is known that, $^n{P_r} = \dfrac{{n!}}{{(n - !)}}$, where $0 \leqslant r \leqslant n$

$\therefore 0 \leqslant r \leqslant n$

Hence, $r \ne 9$

$\therefore r = 4$

 

8. How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?

Ans: The number of different letters in the given word is 8.


Thus, the number of words than can be formed without repetition is number of permutations of 8 different objects taken 8 at a time = $^8{{\text{P}}_8}{\mkern 1mu}  = {\mkern 1mu} 8!$


Therefore, number of words formed = 8! = 40320.

 

9. How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated if,

(i) 4 letters are used at a time

(ii) all letters are used at a time

(iii) all letters are used but first letter is a vowel.

Ans: Number of different letters in the given word = 6


(i) Number of 4-letter words that can be formed from the letters of the given word without repetition is permutations of 6 different objects taken 4 at a time.


Therefore, Number of 4 letter words that can be formed

${\text{  = }}{{\text{ }}^{\text{6}}}{{\text{P}}_{\text{4}}}{\mkern 1mu} {\text{  =  }}{\mkern 1mu} \dfrac{{{\text{6!}}}}{{{\text{(6}}{\mkern 1mu} {\text{  -  }}{\mkern 1mu} {\text{4)}}}}{\text{  =  }}\dfrac{{{\text{6!}}}}{{{\text{2!}}}}{\text{  =  }}\dfrac{{{\text{6}} \times {\text{5}} \times {\text{4}} \times {\text{3}} \times {\text{2}}!}}{{{\text{2!}}}} = 6 \times 5 \times {\text{4}} \times {\text{3}}{\mkern 1mu} {\text{  =  }}{\mkern 1mu} {\text{360}}$


(ii) Words that can be formed using all the letters of the given word is permutation of 6 different objects taken 6 at a time

$^6{{\text{P}}_6}{\mkern 1mu}  = {\mkern 1mu} \,6!{\mkern 1mu} $

Therefore, number of words that can be formed is

$ = 6!{\mkern 1mu}  = {\mkern 1mu} 6 \times 5 \times 4 \times 3 \times 2 \times 1{\mkern 1mu}  = {\mkern 1mu} 720$


(iii) There are two different vowels in the word MONDAY which occupies the rightmost place of the words formed. Hence, there are 2 ways.


Since, it is without repetition and the rightmost place is occupied, the remaining five vacant places can be filled by 5 different letters. Hence, 5! Ways.


Therefore, number of words that can be formed = $5!{\mkern 1mu}  \times {\mkern 1mu} 2{\mkern 1mu}  = {\mkern 1mu} 120{\mkern 1mu}  \times 2{\mkern 1mu}  = {\mkern 1mu} 240$

 

10. In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?

Ans: I appears 4 times, S appears 4 times, P appears 2 times and M appears one time in the given word

Thus, number of permutations of the given word  

$ = \dfrac{{11!}}{{4!4!2!}} = \dfrac{{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4!}}{{4! \times 4 \times 3 \times 2 \times 1 \times 2 \times 1}} = \dfrac{{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5}}{{4 \times 3 \times 2 \times 1 \times 2 \times 1}}$


Since, there are 4 I’s in the given word, they can be treated as a single object. This single object with the remaining 7 objects will together be 8 objects.


These 8 objects in which 4 S’s and 2 P’s is arranged in $\dfrac{{8!}}{{4!2!}}$ = 840 ways


Thus, number of ways where I’s occur together = 840


Therefore, number of permutations where I’s do not occur together is =34650 – 840 = 33810

 

11. In how many ways can the letters of the word PERMUTATIONS can be arranged if the

(i) Words start with P and end with S.

(ii) Vowels are all together.

(iii) There are always 4 letters between P and S?

Ans: There are 2 T’s and other letters are repeated only once in the word PERMUTATION.

(i) Here, P and S are fixed at extreme ends (P at left and S at right end).

Hence, 10 letters are left.

Therefore, required number of permutations 

 $\dfrac{{10!}}{{2!}}{\mkern 1mu}  = {\mkern 1mu} 1814400$

(ii) Since, all the 5 vowels appear only once and they should always occur together, they can be treated as a single object.

This single object with the remaining 7 objects will be 8 objects.

Thus, 8 objects in which 2 T’s can be arranged together is 

$\dfrac{{8!}}{{2!}}$ways

Also, the five different vowels can be arranged in 5! ways.

Thus, by multiplication principle, the number of arrangements

$ = \dfrac{{8!}}{{2!}} \times 5!{\mkern 1mu}  = {\mkern 1mu} 2419200$

(iii) The letters can be arranged in such a way that there are 4 letters between P and S.

Thus, letters P and S are fixed. The remaining 10 letters in which 2 T’s is arranged in

$\dfrac{{10!}}{{2!}}$ways.

Also, there are 4 letters between P and S = $2{\mkern 1mu}  \times {\mkern 1mu} 7{\mkern 1mu}  = {\mkern 1mu} 14$ways

Thus, by multiplication principle, the number of arrangements = $\dfrac{{10!}}{{2!}}{\mkern 1mu}  \times {\mkern 1mu} 14{\mkern 1mu}  = {\mkern 1mu} 25401600$


Exercise 6.4

1.If $^{n}{{C}_{8}}{{=}^{n}}{{C}_{2}}$, find $^{n}{{C}_{2}}$.

Ans:Since, we know that $^{n}{{C}_{a}}{{=}^{n}}{{C}_{b}}\Rightarrow a=b\text{ or }m=a+b$

That is,

$^{n}{{C}_{2}}{{=}^{10}}{{C}_{2}}=\frac{10!}{2!(10-2)!}=\frac{10!}{2!8!}=\frac{10\times 9\times 8!}{2\times 1\times 8!}=45$

Hence, $^{n}{{C}_{2}}=45$.


2. Determine n if

(i) $^{2n}{{C}_{3}}:{{\text{ }}^{n}}{{C}_{3}}=12:1$      

(ii) $^{2n}{{C}_{3}}:{{\text{ }}^{n}}{{C}_{3}}=11:1$

Ans: $\frac{^{2n}{{C}_{3}}}{^{n}{{C}_{3}}}=\frac{12}{1}$

 $  \Rightarrow \dfrac{{{\text{(2n)!}}}}{{{\text{(2n}}\,{\text{ - }}\,{\text{3)!3!}}}}\,\times \dfrac{{{\text{3!(n - }}\,{\text{3)!}}}}{{{\text{n!}}}}\,{\text{ = }}\,\dfrac{{{\text{12}}}}{{\text{1}}}  $

 $  \Rightarrow \dfrac{{{\text{(2n)(2n}}\,{\text{ - }}\,{\text{1)(2n}}\,{\text{ - }}\,{\text{2)(2n}}\,{\text{ - }}\,{\text{3)!}}}}{{{\text{(2n}}\,{\text{ - }}\,{\text{3)!}}}}\,\times \,\dfrac{{{\text{(n}}\,{\text{ - }}\,{\text{3)!}}}}{{{\text{n(n}}\,{\text{ - }}\,{\text{1)(n}}\,{\text{ - }}\,{\text{2)(n}}\,{\text{ - }}\,{\text{3)!}}}}\,{\text{ = }}\,{\text{12}}  $

  $ \Rightarrow \dfrac{{{\text{2(2n}}\,{\text{ - }}\,{\text{1)(2n}}\,{\text{ - }}\,{\text{2)}}}}{{{\text{(n}}\,{\text{ - }}\,{\text{1)(n}}\,{\text{ - }}\,{\text{2)}}}}{\text{ = 12}}  $

  $ \Rightarrow \dfrac{{{\text{4(2n}}\,{\text{ - }}\,{\text{1)(n}}\,{\text{ - }}\,{\text{1)}}}}{{{\text{(n}}\,{\text{ - }}\,{\text{1)(n}}\,{\text{ - }}\,{\text{2)}}}}{\text{ = 12}}  $

  $ \Rightarrow \dfrac{{{\text{(2n}}\,{\text{ - }}\,{\text{1)}}}}{{{\text{(n}}\,{\text{ - }}\,{\text{2)}}}}{\text{ = 3}}  $

  $ \Rightarrow {\text{2n}}\,{\text{ - }}\,{\text{1}}\,{\text{ = }}\,{\text{3(n}}\,{\text{ - }}\,{\text{2)}}  $

  $ \Rightarrow {\text{2n}}\,{\text{ - }}\,{\text{1}}\,{\text{ = }}\,{\text{3n}}\,{\text{ - }}\,{\text{6}}  $

  $ \Rightarrow {\text{3n}}\,{\text{ - }}\,{\text{2n}}\,{\text{ = }}\,{\text{ - 1}}\,{\text{ + 6}}$

  $ \Rightarrow {\text{n}}\,{\text{ = }}\,{\text{5}}  $ 

${\text{(ii) }}\frac{^{2n}{{C}_{3}}}{^{n}{{C}_{3}}}=\frac{11}{1}  $

 $  \Rightarrow \dfrac{{{\text{(2n)!}}}}{{{\text{3!(2n}}\,{\text{ - }}\,{\text{3)!}}}} \times \dfrac{{{\text{3!(n}}\,{\text{ - }}\,{\text{3)!}}}}{{{\text{n!}}}}\,{\text{ = }}\,{\text{11}}  $

  $ \Rightarrow \dfrac{{{\text{(2n)(2n}}\,{\text{ - }}\,{\text{1)(2n}}\,{\text{ - }}\,{\text{2)(2n}}\,{\text{ - }}\,{\text{3)!}}}}{{{\text{(2n}}\,{\text{ - }}\,{\text{3)!}}}}\, \times \,\dfrac{{{\text{(n}}\,{\text{ - }}\,{\text{3)!}}}}{{{\text{n(n}}\,{\text{ - }}\,{\text{1)(n}}\,{\text{ - }}\,{\text{2)(n}}\,{\text{ - }}\,{\text{3)!}}}}  $

 $  \Rightarrow \dfrac{{{\text{2(2n}}\,{\text{ - }}\,{\text{1)(2n}}\,{\text{ - }}\,{\text{2)}}}}{{{\text{(n}}\,{\text{ - }}\,{\text{1)}}\left( {{\text{n}}\,{\text{ - }}\,{\text{2}}} \right)}}{\text{ = 1}}  $

 $  \Rightarrow \dfrac{{{\text{4(2n}}\,{\text{ - }}\,{\text{1)}}}}{{{\text{(n}}\,{\text{ - }}\,{\text{2)}}}}{\text{ = 11}}  $

   $\Rightarrow {\text{4(2n}}\,{\text{ - }}\,{\text{1)}}\,{\text{ = }}\,{\text{11(n}}\,{\text{ - }}\,{\text{2)}}  $

  $ \Rightarrow {\text{8n}}\,{\text{ - }}\,{\text{4}}\,{\text{ = }}\,{\text{11n}}\,{\text{ - }}\,{\text{22}}  $

  $ \Rightarrow {\text{11n}}\,{\text{ - }}\,{\text{8n}}\,{\text{ = }}\,{\text{22}}\,{\text{ - }}\,{\text{4}}  $

  $ \Rightarrow {\text{3n}}\,{\text{ = }}\,{\text{18}}  $

   $\Rightarrow {\text{n}}\,{\text{ = }}\,{\text{6}}  $ 


3. How many chords can be drawn through 21 points on a circle?

Ans: To draw one chord, 2 points are needed.

The number of combinations should be counted to find the number of chords to be drawn through the 21 points on the circle.

Thus, there are combinations of 21 points taken 2 at a time.

Therefore, number of chords is

$^{21}{{C}_{2}}=\frac{21!}{2!(21-2)!}=\frac{21!}{2!19!}=\frac{21\times 20}{2}=210$.


4. In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls.

Ans: Given: 3 boys and 3 girls are to be selected from 5 boys and 4 girls.

Out of 5 boys, 3 boys can be selected in $^{5}{{C}_{3}}$ ways.

Out of 4 girls,  3 girls can be selected in $^{4}{{C}_{3}}$ ways.

Therefore, by multiplication principle, number of ways in which a team of 3 boys and 3 girls can be selected = $^{5}{{C}_{3}}{{\times }^{4}}{{C}_{3}}=\frac{5!}{3!2!}\times \frac{4!}{3!1!}$

                                                = $\frac{5\times 4\times 3!}{3!\times 2}\times \frac{4\times 3!}{3!}=10\times 4$

                                                    = 40


5. Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls each colour.

Ans: Given: Total number of balls: 6 red balls, 5 white balls and 5 blue balls.

3 balls of each colour should be selected making a total of 9 balls.

Therefore,

Out of 6 red balls, 3 balls can be selected in $^{6}{{C}_{3}}$ ways

Out of 5 white balls, 3 balls can be selected in $^{5}{{C}_{3}}$ ways

Out of 5 blue balls, 3 balls can be selected in $^{5}{{C}_{3}}$ ways

Therefore, by multiplication principle, number of ways of selecting 9 balls is =

$^{6}{{C}_{3}}{{\times }^{5}}{{C}_{3}}{{\times }^{5}}{{C}_{3}}=\frac{6!}{3!3!}\times \frac{5!}{3!2!}\times \frac{5!}{3!2!}$

         = $\frac{6\times 5\times 4\times 3!}{3!3\times 2}\times \frac{5\times 4\times 3!}{3!2\times 1}\times \frac{5\times 4\times 3!}{3!2\times 1}$

         = $20\times 10\times 10=2000$


6.Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.

Ans: There are 4 aces in a deck of 52 cards. A combination of 5 cards with one ace has to be made.

One ace is selected in $^{4}{{C}_{3}}$ ways and the remaining 4 cards is selected from the 48 cards in $^{48}{{C}_{4}}$ ways.

Therefore, by multiplication principle, Number of combinations is

$ {{=}^{48}}{{C}_{4}}{{\times }^{4}}{{C}_{1}}=\dfrac{48!}{4!44!}\times \dfrac{4!}{1!3!} $

$ =\dfrac{48\times 47\times 46\times 45\times 44!}{44!4!\times 3\times 2\times 1}\times 4! $

 $ =\dfrac{48\times 47\times 46\times 45}{3\times 2\times 1} $

 $ =778320 $


7. In how many ways can one select a cricket team of 11 from 17 players in which 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers?

Ans: Given: There are 5 bowlers out of 17 players.

A cricket team of 11 is selected with exactly 4 bowlers.

Thus, 4 bowlers can be selected in  $^{5}{{C}_{4}}$ ways and out of remaining 12 players, 7 players are selected in $^{12}{{C}_{7}}$ ways.

Thus, by multiplication principle, required number of ways of selecting cricket team = $^{5}{{C}_{4}}{{\times }^{12}}{{C}_{7}}=\frac{5!}{4!1!}\times \frac{12!}{7!5!}=5!\times \frac{12\times 11\times 10\times 9\times 8}{5!\times 4\times 3\times 2\times 1}=3960$


8. A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.

Ans: There are 5 black and 6 red balls in the bag.2 black balls can be selected out of 5 black balls in $^{5}{{C}_{2}}$ways and 3 red balls can be selected out of 6 red balls in $^{6}{{C}_{3}}$ways.

Thus, by multiplication principle, required number of ways of selecting 2 black and 3 red balls = $^{5}{{C}_{2}}{{\times }^{6}}{{C}_{3}}=\dfrac{5!}{2!3!}\times \dfrac{6!}{3!3!}$

                                  = $\dfrac{5\times 4}{2}\times \dfrac{6\times 5\times 4}{3\times 2\times 1}=10\times 20$                    

  $  = 200$


9.In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?

Ans: Given: Out of 9 courses, 2 courses are compulsory.

Thus, out of the remaining 7 courses, every student has to choose 3 courses. This can be chosen in $^{7}{{C}_{3}}$ ways.

Thus, required number of ways of choosing the programme is: $^{7}{{C}_{3}}=\frac{7!}{3!4!}=\frac{7\times 6\times 5\times 4!}{3\times 2\times 1\times 4!}=35$ 


Miscellaneous Exercise

1. How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER?

Ans:  There are 3 vowels i.e. A, U and E and 5 consonants i.e. D, G, H, T and R in the given word.

Therefore, number of ways of selecting 2 vowels out of 3 = $^{3}{{C}_{2}}$=3

Number of ways of selecting 3 consonants out of 5 = $^{5}{{C}_{3}}=10$

Thus, by multiplication principle, number of combinations = $3\times 10=30$.

Each of these combinations can be arranged in 5! ways.

Hence, number of different words = $30\times 5!=3600$.


2. How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that vowels and consonants occur together?

Ans: There are 5 vowels i.e. A, E, I, O and U and 3 consonants i.e. Q, T and N.

Since, vowels and consonants occur together, both (AEIOU) and (QTN) can be considered as single objects.

Thus, there are 5! Permutations of 5 vowels taken all at a time and 3! permutations of 3 consonants taken all at a time.

Therefore, by multiplication principle, the number of words = $2!\, \times \,5!\, \times \,3!\, = \,1440$.


3. A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of :

(i) exactly 3 girls? (ii) at least 3 girls? (iii) at most 3 girls?

Ans: (i) Out of 9 boys and 4 girls, a committee of 7 has to be formed.

Given: exactly 3 girls should be there in a committee, hence, number of boys = (7 - 3) = 4 boys only.

Therefore, number of ways 

$^{4}{{C}_{3}}{{\times }^{9}}{{C}_{4}}=\frac{4!}{3!1!}\times \frac{9!}{4!5!}$

= $4\times \frac{9\times 8\times 7\times 6\times 5!}{4\times 3\times 2\times 5!}$

= 504

(ii) Given: at least 3 girls are required in each committee. This can be done in 2 ways

(a) 3 girls and 4 boys or (b) 4 girls and 3 boys

3 girls and 4 boys can be selected in  $^{4}{{C}_{3}}{{\times }^{9}}{{C}_{4}}$ ways.

4 girls and 3 boys an be selected in  $^{4}{{C}_{4}}{{\times }^{9}}{{C}_{3}}$ ways.

Thus, number of ways =

$^{4}{{C}_{3}}{{\times }^{9}}{{C}_{4}}{{+}^{4}}{{C}_{4}}{{\times }^{9}}{{C}_{3}}$

= 504+84

= 588

(iii) Given: at most 3 girls in every committee. This can be done in 4 ways

(a) 3 girls and 4 boys (b) 2 girls and 5 boys.

(c) 1 girl and 6 boys (d) No girl and 7 boys

3 girls and 4 boys can be selected in $^{4}{{C}_{3}}{{\times }^{9}}{{C}_{4}}$ ways

2 girls and 5 boys can be selected in $^{4}{{C}_{2}}{{\times }^{9}}{{C}_{5}}$ ways

1 girl and 6 boys can be selected in  $^{4}{{C}_{1}}{{\times }^{9}}{{C}_{6}}$ ways

No girl and 7 boys can be selected in $^{4}{{C}_{0}}{{\times }^{9}}{{C}_{7}}$ ways.

Thus, number of ways

$ {{=}^{4}}{{C}_{3}}{{\times }^{9}}{{C}_{4}}{{+}^{4}}{{C}_{2}}{{\times }^{9}}{{C}_{5}}{{+}^{4}}{{C}_{1}}{{\times }^{9}}{{C}_{6}}{{+}^{4}}{{C}_{0}}{{\times }^{9}}{{C}_{7}} $

 $ =\frac{4!}{3!1!}\times \frac{9!}{4!5!}+\frac{4!}{2!2!}\times \frac{9!}{5!4!}+\frac{4!}{1!3!}\times \frac{9!}{6!3!}+\frac{4!}{0!4!}\times \frac{9!}{7!2!} $ 

 $ =504+756+336+36 $ 

       $=1632 $


4. If the different permutations of all the letter of the word EXAMINATION are listed as in a dictionary, how many words are there in the list before the first word starting with E?

Ans: There are a total of 11 letters out of which A, I and N appears 2 times and other letters appear only once.

The words starting with A will be the words listed before the words starting with E.

Thus, words starting with letter A will have letter A fixed at its extreme left end.

And remaining 10 letters are rearranged all at a time.

In the remaining 10 letters, there are 2 I’s and 2 N’s.

Number of words starting with A = $\dfrac{{10!}}{{2!2!}}\, = \,907200$

Therefore, required number of words is 907200.


5. How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated?

Ans: A number can be divisible by 10 only if its units digit is 0.

Hence, 0 is fixed at units place.

Therefore, the 5 vacant places can be filled by remaining 5 digits.

These 5 empty places can be filled in 5! ways.

Therefore, number of 6-digit numbers = 5! = 120.


6. The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet?

Ans: Given: 2 vowels and 2 consonants should be selected from the English alphabet.

We know that, there are 5 vowels.

Thus, number of ways of selecting 2 vowels out of 5 = \[^{5}{{C}_{2}}=\frac{5!}{2!3!}=10\]

Also we know that, there are 21 consonants.

Thus, number of ways of selecting 2 consonants out of 21 = \[^{21}{{C}_{2}}=\frac{21!}{2!19!}=210\]

Thus, total number of combinations of selecting 2 vowels and 2 consonants is

= $10\, \times \,210\, = \,2100$.

Every 2100 combination consists of 4 letters, which can be arranged in 4! ways.

Therefore, number of words = $2100\, \times \,4!\, = \,50400$


7. In an examination, a paper consists of 12 Qs divided into 2 parts i.e., Part I and Part II, containing 5 and 7 Qs, respectively. A student is required to attempt 8 Qs in all, selecting at least 3 from each part. In how many ways can a student select the Qs?

Ans:  Given: 12 Qs are divided into 2 parts – Part I and Part II consisting of 5 and 7 Qs respectively. A student must attempt 8 Qs with atleast 3 from each part. This can be done as:

  1. 3 Qs from part I and 5 Qs from part II.
  2. 4 Qs from part I and 4 Qs from part II.
  3. 5 Qs from part I and 3 Qs from part II.
The first case can be selected in $^{5}{{C}_{3}}{{\times }^{7}}{{C}_{5}}$ ways.

The second case can be selected in $^{5}{{C}_{4}}{{\times }^{7}}{{C}_{4}}$ ways.

The third case can be selected in $^{5}{{C}_{5}}{{\times }^{7}}{{C}_{3}}$ ways.

Thus, number of ways of selecting Q’s

${{=}^{5}}{{C}_{3}}{{\times }^{7}}{{C}_{5}}{{+}^{5}}{{C}_{4}}{{\times }^{7}}{{C}_{4}}{{+}^{5}}{{C}_{5}}{{\times }^{7}}{{C}_{3}}$ 

 $ =\frac{5!}{2!3!}\times \frac{7!}{2!5!}+\frac{5!}{4!1!}\times \frac{7!}{4!3!}+\frac{5!}{5!0!}\times \frac{7!}{3!4!} $

$ =210+175+35 $

 $=420 $


8. Determine the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.

Ans:  Out of 52 cards, 5-card combinations have to be made in such a way that in each selection of 5 cards, there is exactly one king. We know that in 52 cards, there are 4 kings.

Out of 4 kings, 1 king can be selected in $^{4}{{C}_{1}}$ ways.

Out of the remaining 48 cards, 4 cards can be selected in $^{48}{{C}_{4}}$ ways.

Therefore, number of 5-card combinations =  $^{4}{{C}_{1}}{{\times }^{48}}{{C}_{4}}$ ways.


9. It is required to seat 5 men and 4 women in a row so that the women can occupy the even places. 

How many such arrangements are possible?

Ans: Given: 5 men and 4 women should be seated so that women always occupy the even places.

Thus, the men can be seated in 5! ways.

For each arrangement the women can be seated only in even places.

Thus, the women can be seated in 4! ways.

Therefore, possible number of arrangements = $4!\, \times \,5!\, = \,24\, \times \,120\, = \,2880$


10. From the class of 25 students, 10 are to be chosen for an excursion party. There are 3 students who decide that either all of them will join or none of them will join. In how many ways can the excursion party be chosen?

Ans:  Given: 10 are chosen for an excursion party out of 25 students.

There are 2 cases since 3 students decide either all or one of them will join.

Case 1: All the 3 students join.

The remaining 7 students can be chosen out of 22 students in $^{22}{{C}_{7}}$ ways.

Case 2: None of the 3 students join.

Thus, 10 students can be chosen in $^{22}{{C}_{10}}$ ways.

Hence, number of ways of choosing excursion party = $^{22}{{C}_{7}}{{+}^{22}}{{C}_{10}}$


11. In how many words can the letters of the word ASSASSINATION be arranged so that all the S’s are together?

Ans: There are 3 A’s, 4 S’s, 2 I’s and all other letters appear only once in the word ASSASSINATION.

The given word should be arranged such that all the S’s are together.

The 4 S’s can be treated as a single object for time being. This ingle object with the remaining objects will be 10 objects together.

These 10 objects with 3 A’s, 2 I’s and 2 N’s can be arranged in $\dfrac{{10!}}{{2!3!2!}}$ways.

Therefore, number of ways of arranging the given word = $\dfrac{{10!}}{{2!3!2!}}\, = \,15120$.


Class 11 Maths Chapter 6: Exercises Breakdown

Exercise

Number of Questions

Exercise 6.1

6 Questions & solutions

Exercise 6.2

5 Questions & solutions

Exercise 6.3

11 Questions & solutions

Exercise 6.4

9 Questions & solutions

Miscellaneous Exercise

11 Questions and solutions


Conclusion

In conclusion, Class 11 Permutation and Combination Chapter 6  is a vital part of your Class 11 Maths syllabus. This chapter helps you understand the principles of counting, arrangements, and selections, which are fundamental concepts in probability and various real-life applications.Over the years, questions from this chapter have been frequently asked in previous exams. Typically, you can expect around 3-4 questions from Class 11 Maths Chapter 6 PDF, varying in difficulty from basic to advanced levels. 


Other Study Material for CBSE Class 11 Maths Chapter 6


Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.



Important Related Links for CBSE Class 11 Maths

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FAQs on NCERT Solutions for Class 11 Maths Chapter 6 Permutations And Combinations

1. How do you calculate the factorial of a number (n!)?

Multiply the integer by every positive integer below it down to 1. For example, 5! equals 5 × 4 × 3 × 2 × 1 = 120. Remember the special case: by definition, 0! = 1. To check your work, ensure your product includes all integers from n down to 1.

2. When should you use the Fundamental Principle of Multiplication?

Use this principle to find the total number of outcomes when multiple independent events occur in sequence. If a first event has 'm' possible outcomes and a second has 'n' outcomes, the total number of ways for both to occur is m × n. Simply multiply the options for each step.

3. How can you download the permutations and combinations Class 11 NCERT PDF?

Find the download button for the Class 11 Maths Chapter 6 solutions on this page. Click the link to save the Free PDF to your device. This allows you to study the permutations and combinations class 11 ncert solutions offline, which is great for revision without an internet connection.

4. What is the method for finding permutations when some objects are identical?

First, calculate the factorial of the total number of objects (n!). Then, divide this result by the product of the factorials of the counts for each type of identical object (p₁!, p₂!, etc.). The formula is n! / (p₁! * p₂! * ...), which corrects for overcounting repeated items.

5. How can you use NCERT solutions to quickly check your answers?

Solve a question from your NCERT textbook exercise on your own first. Then, open the solutions and compare your final answer and key formula application. This method helps you instantly spot errors in your logic or calculation without simply copying the provided steps.

6. How do you solve a permutation problem where the order of arrangement matters?

Instruction: Use the permutation formula, P(n, r) = n! / (n-r)!, to find the number of ways to arrange 'r' objects selected from a set of 'n' distinct objects.

Why it matters: This formula is essential when the sequence or position of items is important. Common examples include arranging letters in a word, assigning specific job roles, or determining prize winners.

Steps:

  • Identify the total number of distinct objects available, which is 'n'.
  • Determine the number of objects you need to arrange, which is 'r'.
  • Substitute 'n' and 'r' into the permutation formula.
  • Calculate the factorials and simplify the expression to find the answer.

Check: Ensure that n ≥ r. The result must always be a positive integer. A common mistake is using this formula for selection problems where order does not matter (which requires combinations).

Solving permutation problems involves identifying n and r, applying the P(n, r) formula, and simplifying the factorials.

7. How can the Class 11 Maths Chapter 6 NCERT Solutions be used for self-assessment?

Instruction: Our NCERT Solutions for Class 11 Maths Chapter 6 provide a reliable way to gauge your understanding by comparing your independent work against expert-verified methods.

Why it matters: Self-assessment helps you pinpoint weak areas before exams. It shifts you from passive learning to active problem-solving, which is crucial for building confidence and improving your scores.

Steps:

  • Attempt a full exercise from the NCERT textbook without any help.
  • After finishing, open the Vedantu solutions page for that specific exercise.
  • Compare your final answers first. If an answer is incorrect, review the step-by-step logic.
  • Note down any conceptual errors, like using permutation instead of combination.

Tip: Do not just look at the final answer. Focus on understanding the method and the formulas used in each step, as this is where marks are often awarded in exams.

8. What is the method for solving combination problems where selection order is irrelevant?

Instruction: Apply the combination formula, C(n, r) = n! / [r!(n-r)!], to calculate the number of ways to choose 'r' objects from a set of 'n' distinct objects.

Why it matters: This is used for scenarios like selecting a committee from a group of people or choosing a set of books, where the composition of the group is what counts, not the sequence of selection.

Steps:

  • Confirm the problem is about selection, not arrangement. Keywords like “choose” or “select” are good indicators.
  • Identify the total number of items available (n).
  • Determine the number of items to choose (r).
  • Substitute these values into the C(n, r) formula and simplify.

Check: The number of combinations C(n, r) will always be less than or equal to the number of permutations P(n, r) for the same n and r, as combinations do not count different orderings.

For combination problems, use the C(n, r) formula when order does not matter, focusing on selection rather than arrangement.

9. How do you decide whether to use permutation or combination for a question?

Instruction: Analyse the problem statement for keywords that imply order is important. If the arrangement matters, use permutation. If only the selection of a group matters, use combination.

Why it matters: Correctly identifying the problem type is the most critical first step. Choosing the wrong formula will lead to a completely incorrect answer for permutations and combinations questions and answers class 11.

Steps:

  • Read the question carefully to understand the core task.
  • Look for words like "arrange," "order," "rank," or "position." These signal a permutation.
  • Look for words like "select," "choose," "group," or "committee." These signal a combination.
  • Ask: "Does changing the order of the chosen items create a new outcome?" If yes, it's a permutation. If no, it's a combination.

Example: Arranging 3 people in a line is a permutation. Choosing a team of 3 people is a combination.

10. What is the best way to practise the Miscellaneous Exercise for Permutations and Combinations Class 11?

Instruction: Tackle the Miscellaneous Exercise questions after you have thoroughly completed all the main exercises in the chapter. Use the solutions PDF as a final check, not a guide.

Why it matters: The Miscellaneous Exercise contains more complex problems that mix concepts from the entire chapter. Mastering these is a strong indicator of exam readiness.

Steps:

  • Revise all chapter concepts and formulas before starting.
  • Attempt each problem on your own, trying to identify the correct approach (P, C, or mixed).
  • If you are stuck, refer to the solution to understand only the initial hint.
  • After solving, compare your full method with the one in the Free PDF to learn the most efficient way.

Tip: Many of these questions require a combination of the Fundamental Principle of Counting along with permutation or combination formulas. This tests your deeper understanding of the chapter.