Chapter 6. Permutations and Combinations

Class 11 Maths Chapter 6. Permutations and Combinations Exercise 6.1 Solutions :

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?

Solution :  (i) Repetition of digits is allowed:

[ In this case, for each of the three positions in the 3-digit number, we have 5 choices (1, 2, 3, 4, or 5). ]

Using Fundamental principle of counting , we have

Total number of 3-digit numbers = 5 (choices for the first digit) × 5 (choices for the second digit) × 5 (choices for the third digit)

(ii) Repetition of digits is not allowed :

[ In this case, for the first digit, we have 5 choices (1, 2, 3, 4, or 5).

Once we have chosen the first digit, there are only 4 remaining choices for the second digit (since we can't repeat the first digit), and for the third digit, there are 3 remaining choices. ]

Using Fundamental principle of counting , we have

Total number of 3-digit numbers = 5 (choices for the first digit) × 4 (choices for the second digit) × 3 (choices for the third digit) = 5 × 4 × 3 = 60

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

Solution :  The number of 3-digit even numbers that can be formed from the digits 1, 2, 3, 4, 5, and 6 with repetition allowed .

[ A 3-digit even number will end in 2, 4, or 6. The first two digits can be any of the six given digits.

For the last digit, there are 3 choices (2, 4, or 6).

For the first two digits, there are 6 choices for each digit. ]

Using Fundamental principle of counting , we have

Total number of 3-digit even numbers = 6 (choices for the first digit) × 6 (choices for the second digit) × 3 (choices for the third digit) = 6 × 6 × 3 = 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?

Solution :  The number of 4-letter codes that can be formed using the first 10 letters of the English alphabet with no letter repeated .

[ There are 10 choices for the first letter, 9 choices for the second letter (since one letter has already been used), 8 choices for the third letter, and 7 choices for the fourth letter.]

Using Fundamental principle of counting , we have

Number of 4-letter codes = 10 (choices for the first letter) × 9 (choices for the second letter) × 8 (choices for the third letter) × 7 (choices for the fourth letter) = 10 × 9 × 8 × 7 = 5,040.

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?

Solution:  The number of 5-digit telephone numbers that can be constructed using the digits 0 to 9, with the condition that each number starts with "67" and no digit appears more than once .

[ We have already used two digits for "67" at the beginning.

Now, we have eight digits (0 to 9, excluding 6 and 7) to choose from for the remaining three positions. Since no digit can be repeated, for the first remaining digit, we have 8 choices. For the second remaining digit, we have 7 choices, and for the third remaining digit, we have 6 choices. ]

Using Fundamental principle of counting , we have

Total number of 5-digit telephone numbers = 8 (choices for the third digit) × 7 (choices for the fourth digit) × 6 (choices for the fifth digit) = 8 × 7 × 6 = 336.

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

Solution :  When a coin is tossed 3 times, each toss has 2 possible outcomes: either it lands heads (H) or tails (T). [ Since the outcomes of each toss are independent of one another .]

Using Fundamental principle of counting , we have

Total number of possible outcomes = 2 (possibilities for the first toss) × 2 (possibilities for the second toss) × 2 (possibilities for the third toss)   .

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 ?

Solution :  Using the fundamental principle of counting, we have

[ For each flag in the signal, we have 5 choices (since there are 5 different flags). When we stack one flag below the other, we are essentially creating a pair of flags. ]

Using the fundamental principle of counting, we have

Number of different signals = 5 (choices for the first flag) × 4 (choices for the second flag) = 5 × 4 = 20.

Class 11 Maths Chapter 7. Permutations and Combinations Exercise 7.2 Solutions :

1. Evaluate  :    (i)                 (ii)  

Solution: (i) We have,                

(ii) We have, 

2. Is  ?        

Solution : LHS :

RHS : 

LHS ≠ RHS

This is not correct .

3. Compute

Solution:  We have, 

4. If , find  .

Solution : We have,  

Thereforefore, the value of  is 64 .
5. Evaluate : , when
(i)      (ii)  .

Solution:  (i) Here,     

We have, 

(ii)  Here,   .

We have, 

Class 11 Maths Chapter 7. Permutations and Combinations Exercise 7. 3 Solutions :

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

Solution : We want to arrange 3 digits out of 9.

The number of 3-digit numbers that can be formed using the digits 1 to 9 with no repetition

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

Solution :  The number of 3-digit even numbers using the digits 1, 2, 3, 4, 6, 7 with no repetition.

[ We want to arrange 3 digits out of 6 (1, 2, 3, 4, 6, 7) to form a 3-digit number.

 Since it's an even number, the last digit must be even (2, 4, 6).

So, for the last digit, we have 3 choices (2, 4, 6).

For the first digit, we have 5 choices (excluding the chosen last digit).

For the second digit, we have 4 choices (excluding the two chosen digits), and for the third digit, We have 3 choices. ]

The number of ways

Now, since half of the 3-digit numbers will be even, then we divided by 2 .

The total number of ways

So, there are 60 different 3-digit even numbers that can be formed using the digits 1, 2, 3, 4, 6, 7 with no repetition.

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?

Solution :  The number of 4-digit numbers using the digits 1, 2, 3, 4, 5 without repetition:

For the first digit :  5 choices.

For the second digit :  4 choices (since you can't repeat the digit from the first position).

For the third digit : 3 choices.

For the fourth digit :  2 choices.

Using the Fundamental Principle of Counting , we have

The total number of 4-digit numbers can be formed = 5 × 4 × 3 × 2 = 120  

So, there are 120 different 4-digit numbers that can be formed using the given digits without repetition.

Again, the number of even 4-digit numbers, the last digit must be even (either 2 or 4).

For the last digit (even) : 2 choices.

For the first digit : 4 choices.

For the second digit : 3 choices.

For the third digit : 2 choices.

Using the Fundamental Principle of Counting , we have

The total number of 4-digit even numbers can be formed  = 2 × 4 × 3 × 2 = 48

So, there are 48 different 4-digit even numbers that can be formed using the given digits without repetition.

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

Solution:  To choose a chairman and a vice chairman from a committee of 8 persons where one person cannot hold more than one position .

For the chairman : 8 choices.

For the vice chairman : 7 choices (since the person chosen as chairman cannot be chosen again).

Using the Fundamental Principle of Counting , we have

The total number of ways can we choose a chairman and a vice chairman = 8 × 7 = 56 .

So, there are 56 ways to choose a chairman and a vice chairman from the committee of 8 persons.

6. Find  if   . 

Solution :  We have,  

Therefore, the value of n is 9 .

7. Find  if  (i)    (ii).

Solution :  (i) We have,    

  or  

     or  (0 ≤ r ≤ 5)

Therefore, the value of   is 3 .

(ii) We have,    

 or 

 or  (0 ≤ r ≤ 5)

Therefore, the value of  is 4 .
8. How many words, with or without meaning, can be formed using all the letters ofthe word EQUATION, using each letter exactly once?

Solution :  The number of words that can be formed using all the letters of the word EQUATION, where each letter is used exactly once . The word EQUATION has 8 letters .

The total number of ways to arrange these letters  

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?

Solution:  (i) If 4 letters are used at a time:

For the first letter : 6 choices (M, O, N, D, A, Y).

For the second letter : 5 choices.

For the third letter : 4 choices.

For the fourth letter : 3 choices.

The number of ways to arrange the letters

So, there are 360 words with 4 letters selected from the word "MONDAY."

(ii) If all letters are used at a time:

The word "MONDAY" has 6 distinct letters .

The number of ways to arrange the letters

So, there are 720 words with all letters used at a time.

(iii) If all letters are used, but the first letter is a vowel:

The word "MONDAY" has 2 vowels (O and A) and 4 consonants (M, N, D, Y).

For the first vowel (O or A) : 2 choices.

For the second letter : 5 choices (excluding the chosen vowel).

For the third letter : 4 choices.

For the fourth letter : 3 choices.

For the fifth letter : 2 choices.

For the sixth letter : 1 choice.

The number of ways to arrange the letters

 The total number of ways to arrange the letters  

So, there are 240 words with all letters used, but the first letter is a vowel .

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

Solution:  The number of distinct permutations of the letters in "MISSISSIPPI" where the four I's do not come together .

The word "MISSISSIPPI" has a total of 11 letters, with the following repetitions:

M : 1 time

I: 4 times

S: 4 times

P: 2 times

The total number of permutations of the letters in "MISSISSIPPI"

Again , the number of permutations where the four I's come together. [Treat the four I's as a single unit] So, we have 8 units in total (1 for "IIII," and 7 for the other letters).

The number of permutations where the four I's come together

The total number of permutations to get the number of permutations where the four I's do not come together = 34650 – 840 = 33810 .

11. In how many ways can the letters of the word PERMUTATIONS 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 ?

Solution : The total number of letters = 12 .

(i) If the word starts with P and ends with S:

There are 10 letters left to arrange (excluding P and S) .

The number of ways in which words can start with P and end with S

(ii) If vowels are all together:

The vowels in the word PERMUTATIONS are E, U, A, I, O.

 Treat these five vowels as a single entity. Now, we have 8 entities in total (P , (EUAIO) , R , M , T , T , N , S) and 5 vowels can be arranged among themselves in 5! ways.

The total number of arrangements

(iii) If there are always 4 letters between P and S :

 We can consider the PS pair as a single entity. Then, we have 10 entities (PS, E, R, M, U, T, A, T, I, O, N) to arrange. The PS pair can be arranged in 2! ways (P and S or S and P).

The number of arrangements

 There are always 4 letters between P and S can filled in 7 ways . So, P and S or S and P can be filled in 7+7 = 14 ways .

The total number of ways the letters of the word PERMUTATIONS can be arranged

Class 11 Maths Chapter 7. Permutations and Combinations Exercise 7.4 Solutions :

1. If , find .

Solution :  We have ,

     [   i.e.,   ]

2. Determine  if
(i)    

(ii)   

Solution : (i) We have,   

Therefore, the value of n is 5 .

(ii) We have, 

Therefore, the value of n is 6 .

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

Solution :  The number of chords that can be drawn through  points on a circle using combinations is given by the combination formula:

 

Here ,  

 

There are 210 chords that can be drawn through 21 points on a circle.

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

Solution :  The number of ways to select a team of 3 boys and 3 girls from 5 boys and 4 girls .

We want to select 3 boys out of 5 in ways and 3 girls out of 4 in ways .

Using fundamental principle of counting , we have

Total number of team selected

So, there are 40 ways to select a team of 3 boys and 3 girls from 5 boys and 4 girls .

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 of each colour.

Solution : The number of ways of selecting 9 balls from 6 red balls, 5 white balls, and 5 blue balls, where each selection consists of 3 balls of each color .

We want to select 3 red balls out of 9 balls in ways , 5 white balls out of 9 balls in ways and 5 blue balls out of 9 balls in ways.

Using fundamental principle of counting , we have

Total number of ways for selecting balls

So, there are 2000 ways to select 9 balls, with 3 balls of each color, from 6 red balls, 5 white balls, and 5 blue balls.

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

Solution : The number of 5-card combinations out of a deck of 52 cards .

We  want to choose one ace out of the four aces and then choose four other non-ace cards from the remaining 48 cards .

  Using fundamental principle of counting , we have

Total number of ways

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

Solution : The number of ways to select a cricket team of eleven from 17 players, where only 5 players can bowl and each team must include exactly 4 bowlers .

We want to choose 4 bowlers out of the 5 who can bowl, and then choose 7 players from the remaining 12 players who are not bowlers.

Using fundamental principle of counting , we have

Total number of ways selecting of a cricket team for eleven players

So, there are 3960 ways to select a cricket team of eleven with exactly 4 bowlers from 17 players .

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.

Solution :   The number of ways to select 2 black and 3 red balls from a bag containing 5 black and 6 red balls.

 We want to choose 2 black balls out of 5 balls in ways and then choose 3 red balls out of 6 in  .

Using fundamental principle of counting , we have

Total number of ways selecting of a cricket team for eleven players  

So, there are 200 ways to select 2 black and 3 red balls from a bag containing 5 black and 6 red balls.

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 ?

Solution :  The number of ways a student can choose a program of 5 courses from 9 courses where 2 specific courses are compulsory .

We want to choose 3 courses from the remaining 7 courses (since 2 are already compulsory).

The total number of  the student to choose a program

So, there are 35 ways for a student to choose a program of 5 courses from 9 courses where 2 specific courses are compulsory .

Class 11 Maths Chapter 7. Permutations and Combinations Miscellaneous Exercise Solutions :

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 ?

Solution :  The number of words, with or without meaning, that can be formed using 2 vowels and 3 consonants from the letters of the word DAUGHTER .

In the word DAUGHTER, there are 3 vowels (A, U, E) and 5 consonants (D, G, H, T, R).

We need to select 2 vowels out of the 3 available vowels in ways and 3 consonants out of the 5 available consonants in ways .

And these 5 selected letters (2 vowels and 3 consonants) to form a word and the words arrange 5! Ways .

Using Fundamental principle of counting , we have

Total number of words

So, there are 3600 different words, with or without meaning, that can be formed using 2 vowels and 3 consonants from the letters of the word DAUGHTER.

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

Solution : The number of words, with or without meaning, that can be formed using all the letters of the word EQUATION at a time, such that the vowels and consonants occur together .

[ We can treat the vowels (E, U, A, I, O) as one group and the consonants (Q, T, N) as another group. This way, we have 2 groups to arrange. ]

In the group of vowels (E, U, A, I, O), we can arrange  5! ways .

In the group of consonants (Q, T, N), we can arrange 3! ways .

We have 2 groups (vowels and consonants), so we can arrange 2! ways .

Using Fundamental principle of counting , we have

The total number of words

So, there are 1440 different words, with or without meaning, that can be formed using all the letters of the word EQUATION, such that the vowels and consonants occur together .

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) atleast 3 girls ? (iii) atmost 3 girls ?

Solution:  The number of ways to form a committee of 7 members from 9 boys and 4 girls .

 (i) Exactly 3 girls:

We need to choose 3 girls from the 4 available girls in ways  and 4 boys from the 9 available boys in ways .

Using Fundamental principle of counting , we have

The total number of ways

(ii) At least 3 girls: Total number of members = 9 + 4 = 13 .

We choose 7 members from 13 members in  ways .

Total number of ways to form the committee

Number of ways with fewer than 3 girls = (choose 0 girls from 4 in ways and 7 boys from 9 in ways ) + (choose 1 girl from 4 in ways and 6 boys from 9 in ways) + (choose 2 girls from 4 in ways and 5 boys from 9 in ways)

Number of ways with fewer than 3 girls

Number of ways with at least 3 girls = Total number of ways  – Number of ways with fewer than 3 girls

Number of ways with at least 3 girls = 1716 – 1128 = 588  

 (iii) At most 3 girls:

We choose 7 members from 13 in  ways .

Total number of ways to form the committee

The number of ways with more than 3 girls = (choose all 4 girls in ways and 3 boys from 9 in ways)

Number of ways with more than 3 girls

Number of ways with at most 3 girls = Total number of ways  –  Number of ways with more than 3 girls

Number of ways with at most 3 girls = 1716 – 84  = 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 this list before the first word starting with E ?

Solution: The number of letter of the word EXAMINATION = 11 .

The permutations that start with 'E', we can consider the remaining 10 letters, as the first 'E' is fixed. Among these 10 letters, there are 2 'I's and 2 'A's.

The number of permutations that start with 'E' 

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 ?

Solution :  A 6-digit number is divisible by 10 if and only if it ends with a 0.

Therefore, the last digit of the 6-digit number must be 0.

We can choose the last digit in only one way, which is by selecting 0.

We need to arrange the remaining 5 digits (1, 3, 5, 7, and 9) in the first five positions without repetition.

There are 5 choices for the first digit (1, 3, 5, 7, or 9), 4 choices for the second digit (one of the remaining four digits), 3 choices for the third digit, 2 choices for the fourth digit, and 1 choice for the fifth digit.

Using the fundamental principle of counting , we have

The number of ways =  (choices for the first digit) × (choices for the second digit) × (choices for the third digit) ×  (choices for the fourth digit) ×  (choice for the fifth digit)

The number of ways = 5 × 4 × 3 × 2 × 1 = 120 ways .

So, there are 120 different 6-digit numbers that can be formed from the digits 0, 1, 3, 5, 7, and 9, which are divisible by 10 and have no repeated digits.

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 ?

Solution : There are 2 vowels out of 5 vowels can be selected in ways .

There are 2 consonants out of 21 can be selected in  ways .

There are 4 positions (2 for vowels and 2 for consonants), and we can arrange them in 4! ways.

Using the fundamental principle of counting, we have

Total number of words

So, there are 50,400 different words with two different vowels and two different consonants that can be formed from the English alphabet.

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

Solution :   Selecting 3 questions from Part I and 5 questions from Part II.

For Part I :  3 questions out of the 5 available in ways.

For Part II : 5 questions out of the 7 available in  ways.

The total number of ways   

Selecting 4 questions from Part I and 4 questions from Part II.

For Part I : 4 questions out of the 5 available in ways.

For Part II : 4 questions out of the 7 available in ways.

The total number of ways  .

Selecting 5 questions from Part I and 3 questions from Part II.

For Part I : 5 questions out of the 5 available in ways.

For Part II : 3 questions out of the 7 available in ways.

The total number of ways   .

Total number of ways

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.

Solution :  There are four kings in a standard deck of 52 cards, one for each suit (hearts, diamonds, clubs, spades).

1King out of 4 king can be selected in  ways .

To select the four non-king cards, we have 48 cards remaining in the deck (52 cards - 4 kings).

The number of ways to choose 4 non-king cards from .

Total combinations

9. It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible ?

Solution : First, arrange the 5 men in the odd positions (positions 1, 3, 5, 7, and 9).

There are 5! ways to arrange the men in these positions.

Next, arrange the 4 women in the even positions (positions 2, 4, 6, and 8).

There are 4! ways to arrange the women in these positions.

Using fundamental  principle counting , we have

Total arrangements = (Number of ways to arrange men) × (Number of ways to arrange women)

= 5! × 4! = 120 × 24 = 2880.

10. From a 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 ?

Solution :  3 students decide to join the excursion party :

In this case, we select 7 more students from the remaining 25 – 3 = 22 students in   ways .

3 students decide not to join the excursion party :

 In this case, we select 10 students from the remaining 25 – 3 = 22 students in  ways .  

Total number of ways  .

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

Solution :   In the word "ASSASSINATION," there are 13 letters in total.

 The letter 'A' appears :  3 times,

              'S' appears  : 4 times,

               'I' appears  : 2 times,

              'N' appears :  2 times,

              ‘O’ appear : 1 time ,

    and 'T' appears : 1 time.

Therefore, the total number of "letters" to arrange is 10 (since "SSSS" as one).

The total number of arrangements =

Therefore, the total number of ways the letters of the word "ASSASSINATION" can be arranged with all the S's together is 151200 ways .