NCERT Solutions Class 11 Maths Chapter 6 Permutation & Combination

Consider LHS 3! + 4!

Computing the left-hand side, we get

3! + 4! = (3 × 2 × 1) + (4 × 3 × 2 × 1)

= 6 + 24

= 30

Again, considering RHS and computing, we get

7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040

Therefore, LHS ≠ RHS

Therefore, 3! + 4! ≠ 7!

 

Given A coin is tossed 3 times, and the outcomes are recorded.

The possible outcomes after a coin toss are head and tail.

The number of possible outcomes at each coin toss is 2.

∴ The total number of possible outcomes after 3 times = 2 × 2 × 2 = 8

 

Given 5 flags of different colours.

We know the signal requires 2 flags.

The number of flags possible for the upper flag is 5.

Now, as one of the flags is taken, the number of flags remaining for the lower flag in the signal is 4.

The number of ways in which signal can be given = 5 × 4 = 20

 

(i) Consider 8!

We know that 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1

= 40320

(ii) Consider 4!-3!

4!-3! = (4 × 3!) – 3!

The above equation can be written as

= 3! (4-1)

= 3 × 2 × 1 × 3

= 18

 

(i) Let the 3-digit number be ABC, where C is at the units place, B at the tens place and A at the hundreds place.

Now, when repetition is allowed,

The number of digits possible at C is 5. As repetition is allowed, the number of digits possible at B and A is also 5 at each.

Hence, the total number possible 3-digit numbers =5 × 5 × 5 =125

(ii) Let the 3-digit number be ABC, where C is at the units place, B at the tens place and A at the hundreds place.

Now, when repetition is not allowed,

The number of digits possible at C is 5. Suppose one of 5 digits occupies place C; now, as the repletion is not allowed, the possible digits for place B are 4, and similarly, there are only 3 possible digits for place A.

Therefore, the total number of possible 3-digit numbers=5 × 4 × 3=60

 

Let the 3-digit number be ABC, where C is at the unit’s place, B at the tens place and A at the hundreds place.

As the number has to be even, the digits possible at C are 2 or 4 or 6. That is, the number of possible digits at C is 3.

Now, as repetition is allowed, the digits possible at B is 6. Similarly, at A, also, the number of digits possible is 6.

Therefore, The total number of possible 3-digit numbers = 6 × 6 × 3 = 108

 

Let the 4-letter code be 1234.

In the first place, the number of letters possible is 10.

Suppose any 1 of the ten occupies place 1.

Now, as repetition is not allowed, the number of letters possible at place 2 is 9. Now, at 1 and 2, any 2 of the 10 alphabets have been taken. The number of alphabets left for place 3 is 8, and similarly, the number of alphabets possible at 4 is 7.

Therefore, the total number of 4-letter codes=10 × 9 × 8 × 7=5040

 

Let the five-digit number be ABCDE. Given that the first 2 digits of each number are 67. Therefore, the number is 67CDE.

As repetition is not allowed and 6 and 7 are already taken, the digits available for place C are 0,1,2,3,4,5,8,9. The number of possible digits at place C is 8. Suppose one of them is taken at C; now the digits possible at place D is 7. And similarly, at E, the possible digits are 6.

∴ The total five-digit numbers with given conditions = 8 × 7 × 6 = 336

 

To find the four-digit number (digits do not repeat),

We will have 4 places where 4-digits are to be put.

So, at the thousand’s place = There are 9 ways as 0 cannot be at the thousand’s place = 9 ways

At the hundredth’s place = There are 9 digits to be filled as 1 digit is already taken = 9 ways

At the ten’s place = There are now 8 digits to be filled as 2 digits are already taken = 8 ways

At unit’s place = There are 7 digits that can be filled = 7 ways

The total number of ways to fill the four places = 9 × 9 × 8 × 7 = 4536 ways

So, a total of 4536 four-digit numbers can be there with no digits repeated.

 

An even number means that the last digit should be even.

The number of possible digits at one’s place = 3 (2, 4 and 6)

⇒ Number of permutations=

One of the digits is taken at one’s place; the number of possible digits available = 5

⇒ Number of permutations=

Therefore, the total number of permutations =3 × 20=60

 

We have a deck of cards that has 4 kings.

The numbers of remaining cards are 52.

Ways of selecting a king from the deck = 4C1

Ways of selecting the remaining 4 cards from 48 cards= 48C4

The total number of selecting the 5 cards having one king always

= 4C1 × 48C4

 

Total number of digits possible for choosing = 5

Number of places for which a digit has to be taken = 4

As there is no repetition allowed,

⇒ Number of permutations =

The number will be even when 2 and 4 are in one’s place.

The possibility of (2, 4) at one’s place = 2/5 = 0.4

The total number of even numbers = 120 × 0.4 = 48

 

Total number of people in committee = 8

Number of positions to be filled = 2

⇒ Number of permutations =

 

Total number of different letters in EQUATION = 8

Number of letters to be used to form a word = 8

⇒ Number of permutations =

 

(i) Number of letters to be used =4

⇒ Number of permutations =

(ii) Number of letters to be used = 6

⇒ Number of permutations =

(iii) Number of vowels in MONDAY = 2 (O and A)

⇒ Number of permutations in vowel =

Now, the remaining places = 5

Remaining letters to be used =5

⇒ Number of permutations =

Therefore, the total number of permutations = 2 × 120 =240

 

Total number of letters in MISSISSIPPI =11

Letter Number of occurrence

⇒ Number of permutations =

We take that 4 I’s come together, and they are treated as 1 letter,

∴ Total number of letters=11 – 4 + 1 = 8

⇒ Number of permutations =

Therefore, total number of permutations where four I’s don’t come together = 34650-840=33810

 

(i) Total number of letters in PERMUTATIONS =12

The only repeated letter is T; 2times

The first and last letters of the word are fixed as P and S, respectively.

Number of letters remaining =12 – 2 = 10

⇒ Number of permutations =

(ii) Number of vowels in PERMUTATIONS = 5 (E, U, A, I, O)

Now, we consider all the vowels together as one.

Number of permutations of vowels = 120

Now, the total number of letters = 12 – 5 + 1= 8

⇒ Number of permutations =

Therefore, the total number of permutations = 120 × 20160 = 2419200

(iii) The number of places is as 1 2 3 4 5 6 7 8 9 10 11 12

There should always be 4 letters between P and S.

Possible places of P and S are 1 and 6, 2 and 7, 3 and 8, 4 and 9, 5 and 10, 6 and 11, 7 and 12

Possible ways =7,

Also, P and S can be interchanged,

No. of permutations =2 × 7 =14

The remaining 10 places can be filled with 10 remaining letters,

∴ No. of permutations =

Therefore, the total number of permutations = 14 × 1814400 =25401600

 

Simplifying and computing

⇒ 4 × (2n – 1) = 12 × (n – 2)

⇒ 8n – 4 = 12n – 24

⇒ 12n – 8n = 24 – 4

⇒ 4n = 20

∴ n = 5

⇒ 11n – 8n = 22 – 4

⇒ 3n = 18

∴ n = 6

 

Given 21 points on a circle.

We know that we require two points on the circle to draw a chord.

∴ The number of chords is are

⇒ 21C2=

∴ The total number of chords that can be drawn is 210

 

Given 5 boys and 4 girls in total.

We can select 3 boys from 5 boys in 5C3 ways.

Similarly, we can select 3 boys from 54 girls in 4C3 ways.

∴ The number of ways a team of 3 boys and 3 girls can be selected is 5C3 × 4C3

⇒ 5C3 × 4C3 =

⇒ 5C3 × 4C3 = 10 × 4 = 40

∴ The number of ways a team of 3 boys and 3 girls can be selected is 5C3 × 4C3 = 40 ways

 

Given 6 red balls, 5 white balls and 5 blue balls.

We can select 3 red balls from 6 red balls in 6C3 ways.

Similarly, we can select 3 white balls from 5 white balls in 5C3 ways.

Similarly, we can select 3 blue balls from 5 blue balls in 5C3 ways.

∴ The number of ways of selecting 9 balls is 6C3 ×5C3 × 5C3

∴ 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 is 6C3 ×5C3 × 5C3 = 2000

 

Given a deck of 52 cards.

There are 4 Ace cards in a deck of 52 cards.

According to the question, we need to select 1 Ace card out of the 4 Ace cards.

∴ The number of ways to select 1 Ace from 4 Ace cards is 4C1

⇒ More 4 cards are to be selected now from 48 cards (52 cards – 4 Ace cards)

∴ The number of ways to select 4 cards from 48 cards is 48C4

∴ The number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination is 778320.

 

Given 17 players, in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers.

There are 5 players that can bowl, and we can require 4 bowlers in a team of 11.

∴ The number of ways in which bowlers can be selected is: 5C4

Now, other players left are = 17 – 5(bowlers) = 12

Since we need 11 players in a team and already 4 bowlers have been selected, we need to select 7 more players from 12.

∴ The number of ways we can select these players is: 12C7

∴ The total number of combinations possible is: 5C4 × 12C7

∴ The number of ways we can select a team of 11 players where 4 players are bowlers from 17 players is 3960.

 

Given a bag contains 5 black and 6 red balls

The number of ways we can select 2 black balls from 5 black balls is 5C2

The number of ways we can select 3 red balls from 6 red balls is 6C3

The number of ways 2 black and 3 red balls can be selected is 5C2× 6C3

∴ The number of ways in which 2 black and 3 red balls can be selected from 5 black and 6 red balls is 200.

 

Given 9 courses are available and 2 specific courses are compulsory for every student.

Here, 2 courses are compulsory out of 9 courses, so a student needs to select 5 – 2 = 3 courses

∴ The number of ways in which 3 ways can be selected from 9 – 2(compulsory courses) = 7 are 7C3

∴ The number of ways a student selects 5 courses from 9 courses where 2 specific courses are compulsory is 35.

 

The word DAUGHTER has 3 vowels A, E, and U and 5 consonants D, G, H, T and R.

The three vowels can be chosen in 3C2 as only two vowels are to be chosen.

Similarly, the five consonants can be chosen in 5C3 ways.

∴ The number of choosing 2 vowels and 5 consonants would be 3C2 ×5C3

= 30

∴ The total number of ways of is 30.

Each of these 5 letters can be arranged in 5 ways to form different words = 5P5

Total number of words formed would be = 30 × 120 = 3600

 

In the word EQUATION, there are 5 vowels (A, E, I, O, U) and 3 consonants (Q, T, N).

The numbers of ways in which 5 vowels can be arranged are 5C5

…………… (i)

Similarly, the numbers of ways in which 3 consonants can be arranged are 3P3

…………….. (ii)

There are two ways in which vowels and consonants can appear together.

(AEIOU) (QTN) or (QTN) (AEIOU)

∴ The total number of ways in which vowel and consonant can appear together are 2 × 5C5 × 3C3

∴ 2 × 120 × 6 = 1440

 

(i) Given exactly 3 girls.

The total numbers of girls are 4.

Out of which, 3 are to be chosen.

∴ The number of ways in which choice would be made = 4C3

Numbers of boys are 9 out of which 4 are to be chosen which is given by 9C4

Total ways of forming the committee with exactly three girls.

= 4C3 × 9C4

=

(ii) Given at least 3 girls.

There are two possibilities for making a committee choosing at least 3 girls.

There are 3 girls and 4 boys, or there are 4 girls and 3 boys.

Choosing three girls we have done in (i)

Choosing four girls and 3 boys would be done in 4C4 ways.

And choosing 3 boys would be done in 9C3

Total ways = 4C4 ×9C3

The total number of ways of making the committee are

504 + 84 = 588

(iii) Given at most 3 girls

In this case, the numbers of possibilities are

0 girl and 7 boys

1 girl and 6 boys

2 girls and 5 boys

3 girls and 4 boys

Number of ways to choose 0 girl and 7 boys = 4C0 × 9C7

The number of choosing 3 girls and 4 boys has been done in (1)

= 504

The total number of ways in which a committee can have at most 3 girls are = 36 + 336 + 756 + 504 = 1632

 

In a dictionary, words are listed alphabetically, so to find the words

Listed before E should start with the letter either A, B, C or D.

But the word EXAMINATION doesn’t have B, C or D.

Hence, the words should start with the letter A

The remaining 10 places are to be filled in by the remaining letters of the word EXAMINATION which are E, X, A, M, 2N, T, 2I, 0

Since the letters are repeating, the formula used would be

Where n is the remaining number of letters, p1 and p2 are the number of times the repeated terms occurs.

The number of words in the list before the word starting with E

= words starting with letter A = 907200

 

The number is divisible by 10 if the unit place has 0 in it.

The 6-digit number is to be formed out of which unit place is fixed as 0.

The remaining 5 places can be filled by 1, 3, 5, 7 and 9.

Here, n = 5

And the numbers of choice available are 5.

So, the total ways in which the rest of the places can be filled are 5P5

 

We know that there are 5 vowels and 21 consonants in the English alphabet.

Choosing two vowels out of 5 would be done in 5C2 ways.

Choosing 2 consonants out of 21 can be done in 21C2 ways.

The total number of ways to select 2 vowels and 2 consonants

= 5C2 × 21C2

Each of these four letters can be arranged in four ways 4P4

Total numbers of words that can be formed are

24 × 2100 = 50400

 

The student can choose 3 questions from part I and 5 from part II

Or

4 questions from part I and 4 from part II

5 questions from part 1 and 3 from part II

 

Given there is a total of 9 people.

Women occupy even places, which means they will be sitting in 2nd, 4th, 6th and 8th place where as men will be sitting in 1st, 3rd, 5th,7th and 9th place.

4 women can sit in four places and ways they can be seated= 4P4

5 men can occupy 5 seats in 5 ways.

The number of ways in which these can be seated = 5P5

The total numbers of sitting arrangements possible are

24 × 120 = 2880

 

In this question, we get 2 options, which are

(i) Either all 3 will go

Then, the remaining students in the class are: 25 – 3 = 22

The number of students remained to be chosen for party = 7

Number of ways to choose the remaining 22 students = 22C7

=

(ii) None of them will go

The students going will be 10.

Remaining students eligible for going = 22

The number of ways in which these 10 students can be selected are 22C10

The total number of ways in which students can be chosen is

= 170544 + 646646 = 817190

 

In the given word ASSASSINATION, there are 4 ‘S’. Since all the 4 ‘S’ have to be arranged together, let us take them as one unit.

The remaining letters are= 3 ‘A’, 2 ‘I’, 2 ‘N’, T

The number of letters to be arranged is 9 (including 4 ‘S’).

Using the formula
where n is the number of terms and p1, p2 p3 are the number of times the repeating letters repeat themselves.

Here, p1= 3, p2= 2, p3 = 2

Putting the values in formula we get