NCERT Solutions for Class 11 Maths Chapter 7 - Permutations And Combinations

Exercise 7.1: 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 7.2: 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 7.3: 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 7.4: 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.

Overall, the exercises in NCERT Solutions for Class 11 Maths Chapter 7 – Permutations and Combinations are designed to help students understand and apply the concepts of permutations and combinations in various scenarios. The solutions to these exercises are provided in the textbook, which will help students to check their answers and understand the concepts better.

Access NCERT Solutions Class-11 for Maths Chapter 7 – Permutations and Combinations

Exercise 7.1

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 7.2

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 7.3

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 7.4

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 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 = $\dfrac{{10!}}{{2!3!2!}}\, = \,15120$.

NCERT Solutions for Class 11 Maths Chapter 7 - Permutations and Combinations

Chapter 7 Maths Class 11

Permutation and Combination Class 11 can be defined as the different ways by which elements of a set can be chosen without any replacement and create subsets. This subset selection is termed as permutation, keeping in mind the factor of the order of selection. For combination, the order is not considered as a factor.

Topics Covered in NCERT Solutions Class 11 Math Chapter 7: 

1. 7.1: Introduction

2. 7.2: Fundamental Principle of Counting

3. 7.3: Permutations

• 7.3.1:Permutation when all the objects are distinct

• 7.3.2: Factorial Notation

• 7.3.3: Derivation of the formula nPr

• 7.3.4: Permutations when all the objects are not distinct objects

4. 7.4:  Combinations

What is Permutation and Combination?

• Permutation is the mathematical calculation to determine the number of possible arrangements that can be formed in a set when the order of arrangement matters.

• Combination defines the arrangement of objects where the order in which the objects are selected does not play a crucial role.

Permutation And Combination NCERT Solutions

The entire Permutation and Combination NCERT Solutions are divided into several exercises, and in the following section, we have provided a brief overview of the same -

• Permutation and Combination Class 11: Exercise 7.1

Q. 1 and 2 deals with the formation of 3-digit numbers using the multiplication rule when repetition is allowed, and when not allowed. Permutation and Combination Class 11 Solutions are provided in a precise manner so that you can quickly solve these types of questions.

Q. 3, 4, 5 and 6 are also about the arrangement of numbers by maintaining different conditions specified. Remember that before starting to solve a problem, you must always read the question carefully. The reason is to not miss out any condition, specifically in these types of questions.

• Permutation and Combination Class 11: Exercise 7.2

All the questions of this exercise are related to solving factorials. Few sums also deal with proving L.H.S = R.H.S and substitution. For example, the 2nd question is 3! + 4! = 7! And you need to prove whether the equation is correct or not.

• Permutation and Combination Class 11: Exercise 7.3

This section contains 11 questions in total and focuses primarily on the concept and formulas of permutation. The first two problems are again about the arrangement of the digits, without repetition. Q. 3 and 4 are slightly more complex and require you to provide only the numbers which are odd and even. Q. 5 is similar to problem sum, where you have to find the number of ways in which you can appoint a chairman and vice-chairman, considering each person holds only one position, while Q. 6 and 7 are direct formula based. In Q. 8 and 9, 10 and 11, instead of numbers, you need to solve it for alphabets.

• Permutation and Combination Class 11: Exercise 7.4

Moving on to the next section, you can witness only sums related to combinations. Q. 1, 2 and 3 are simpler ones where you need to find the value of n. In the rest of the questions, you are again supposed to see the number of ways from various combinations. For instance, Q. 4 is about finding a team of girls and boys, Q. 6 is about cards and so on.

• Permutation and Combination Class 11: Miscellaneous exercise

This exercise consists of a variety of Permutation and Combination problems. When starting with the topic, some students may find it challenging to master the same. However, with regular practice and by referring to materials and NCERT Solutions for Class 11 Math Chapter 7 PDF, you can be assured to have a substantial grasp over the topic of Permutation and Combination.

How Class 11 Math NCERT Solutions Chapter 7 PDF Can Help You to Ace the Exams?

Learners can be benefited from the PDF material in the following ways:

• All worked out sums follow a step-by-step process that makes it easier to comprehend.

• The PDF is arranged by proficient subject matter experts and by strictly maintaining the CBSE guidelines as well.

• Students receive a clear idea about the problems or questions that will come in their exams.

Download Class 11 Maths Chapter 7 Solution PDF material today and enhance your problem-solving skills and attain an extra edge over your fellow mates.

Significance of Class 11 Chapters 7 Permutations and Combinations Solutions

This chapter, as mentioned earlier, introduces the topics of permutation and combination for the first time. Hence, students will need the support of the exercise solutions. Understanding the concepts and applying them to solve the problems can only be assessed by attempting solving exercise questions. Hence, the solutions will be a guide to follow and grab hold of the concepts.

The best part of following these solutions is the coverage of the entire syllabus of this chapter. All the exercises are covered in these solutions. The questions have been answered in a precise way by following a stepwise method. This method also contains simpler language that students can easily comprehend and proceed with their preparation.

Following these solutions will help students to overcome any query arising during solving the exercises. There is no need to wait until students can find the precise answers to all the questions included in the exercises.

Hence, you can use these NCERT solutions for Class 11 Permutation and Combination to understand the concepts. Learn more about these topics in this chapter and score more in the exams.

Benefits of NCERT Solutions for Class 11 Maths Permutations and Combinations

Download Class 11 Chapters 7 Permutations and Combinations NCERT Solutions PDF

Get the free PDF version of the solutions for all exercises of this chapter and complete your study material. Check the answers compiled by the subject experts and learn the skills to precisely attempt such questions. Develop your concepts related to permutation and combination and score more in the Class 11 Maths exams.

Conclusion

NCERT Solutions play a crucial role in Class 12 exam prep. Start by thoroughly reading the textbook chapter. After that, solve the NCERT questions for Class 11 Maths Chapter 7 - Permutations and Combinations. You can find detailed solutions on Vedantu, aligning with CBSE guidelines. Download the free NCERT Solutions for Class 11 Chapter 7 “Permutations and Combinations” to guide your exam preparation with expert-reviewed answers.

NCERT Solutions for Class 11 Maths Chapters

• Chapter 1 - Sets

• Chapter 2 - Relations and Functions

• Chapter 3 - Trigonometric Functions

• Chapter 4 - Principle of Mathematical Induction

• Chapter 5 - Complex Numbers and Quadratic Equations

• Chapter 6 - Linear Inequalities

• Chapter 7 - Permutations and Combinations

• Chapter 8 - Binomial Theorem

• Chapter 9 - Sequences and Series

• Chapter 10 - Straight Lines

• Chapter 11 - Conic Sections

• Chapter 12 - Introduction to Three Dimensional Geometry

• Chapter 13 - Limits and Derivatives

• Chapter 14 - Mathematical Reasoning

• Chapter 15 - Statistics

• Chapter 16 - Probability

Along with this, students can also download additional study materials provided by Vedantu, for Chapter 7 of CBSE Class 11 Maths Solutions:

• Chapter 7: Important Questions

• Chapter 7: Important Formulas

• Chapter 7: Revision Notes

• Chapter 7: NCERT Exemplar Solutions

• Chapter 7: RD Sharma Solutions