NCERT Solutions for Class 11 Math Chapter 7 Permutations and Combinations are prepared by the experts at ANAND CLASSES in accordance with the latest syllabus issued by the board.

In this chapter, students learn about the concepts related to Permutation and Combination. Students can easily score full marks in the questions from this chapter by practicing all the exercise questions given in the NCERT textbook. The NCERT Solutions designed by the subject matter experts at ANAND CLASSES cover all the exercise’ questions included in Class 11 Math Chapter 7.

Class 11 Maths Chapter 7 is an interesting part of the syllabus that introduces the concepts of permutation and combination. Students will learn new mathematical principles and topics and apply them to solve the exercise questions. To make this process better, use the NCERT solutions for all the exercises in this chapter. Find out how these problems should be addressed and solved to score more in the exams.

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Exercise 7.1

Q.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) 

The five digits are 1, 2, 3, 4 and 5

As we know that repetition of the digits is allowed, so , unit place can be filled by any of five digits. Similarly , tens and hundreds digits can also be filled by any of five digits.

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) 

The five digits are 1, 2, 3, 4 and 5

As we know that repetition of the digits is not allowed, so, the unit place can be filled by any of five digits.

Tens place can be filled with any of the remaining four digits.

Hundreds place can be filled with any of the remaining three digits.

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

Q.2

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

Solution:

The six digits are 1, 2, 3, 4 ,5 and 6

As we know that repetition of the digits is allowed, so, the unit place can be filled by any of even digits i.e.2, 4 or 6

Similarly, tens and hundreds of digits can also be filled by any of six digits.

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

Q.3

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

Solution:

There are 10 letters of English alphabets.

As we know that repetition of the letters is not allowed, so, the first place can be filled by any of 10 letters.

Second place can be filled with any of the remaining 9 letters.

Third place can be filled with any of the remaining 8 letters.

The fourth place can be filled with any of the remaining 7 letters.

Therefore, Number of 4-letter code can be formed when the repetition of letters is not allowed = 10 × 9 × 8 × 7 = 5040

Hence, 5040 4-letter codes can be formed using the first 10 letters of the English alphabet, if no letter can be repeated.

Q.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:

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

The 10 digits are from 0 to 9.

As we know that repetition of the digits is not allowed, so, the first and second place is filled by two digits 67

Third place C can be filled with any of the remaining 8 digits.

The fourth place D can be filled with any of the remaining 7 digits.

The fifth place E can be filled with any of the remaining 6 digits.

Therefore, Number of 5-digit telephone numbers can be formed when repetition is not allowed  = 8 × 7 × 6 = 336

Q.5

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

Solution:

When a coin is tossed the number of outcomes is 2 i.e. head or tail.

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

Let us denote the ‘Head’ by H and the ‘Tail’ by T. When the coin is tossed 3 times, we have the following Tree-diagram:

When a coin is tossed 3 times then by multiplication principle,

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

Note : The eight possible outcomes are:
HHH, HHT, HTH, HTT, THH, THT, TTH, TTT
(Here HHT is the outcome that the coin turns up ‘Head’ on the first toss, ‘Head’ on the second toss and ‘Tail’ on the third toss whereas HTH is the outcome that the coin turns up ‘Head’ on the first toss, ‘Tail’ on the second toss and ‘Head’ on the third toss. Clearly, HHT and HTH are two different outcomes.)
Note: If a coin is tossed n times, then the number of possible outcomes is 2ⁿ.

Q.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:

Given 5 flags of different colours.

We know the each 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.

Hence, by multiplication principle number of different signals that can be generated = 5 × 4 = 20

Exercise 7.2

Q.1

Evaluate
(i) 8!

(ii) 4! – 3!

Solution:

(i)

Factorial can be given as n! = n × (n-1) × (n-2)………5 × 4 × 3 × 2 × 1

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

Q.2

Is 3! + 4! = 7!?

Solution:

Consider LHS 3! + 4!

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!

Q.3

ComputeSolution:

On expanding, we get

Q.4

If

find x.

Solution:

Multiplying throughout by 8 !, (the largest number factorial in the denominator) we have

Q.5

Evaluate

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

Solution:

Exercise 7.3

Note:

In problems on permutations (⇒ arrangements), both number of things and their order is important. So we must have to apply permutations formulae in the following types of problems:
(i) Words formed by letters (ii) Numbers formed by digits (iii) seating arrangements (iv) signals (v) Letters and Envelopes (vi) tossed (vii) thrown

Q.1

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

Solution:

3-digit numbers have to be formed by using the digits 1 to 9.

Here, the order of digits matters.

There will be as many 3-digit numbers as there are ways of filling 3 vacant places in succession by the 9 given digits.

Therefore, there will be as many 3-digit numbers as there are permutations of 9 different digits taken 3 at a time.

Therefore, the required number of 3-digit numbers = ⁹P₃ = 9 × 8 × 7 = 504 ways.
∴ The required number of 3-digit numbers = 504.

Q.2

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

Solution:

We can use 10 digits 0 to 9.

The number of ways of filling 4 vacant places in succession by the 10 given digits (including 0) is ¹⁰P₄.

But these permutations will include those also where 0 is at the thousand’s place. Such numbers are actually 3-digit numbers (For example 0387) and must be discarded. When 0 is fixed in the thousand’s place, the remaining 9 digits can be arranged in the remaining 3 places in ⁹P₃ ways.
∴ The required number of 4-digit numbers = ¹⁰P₄ – ⁹P₃

= 10 × 9 × 8 × 7 – 9 × 8 × 7 = 5040 – 504 = 4536.

Second mathod

We know that there are 10 digits 0 to 9.
The thousand place can be filled in 9 ways (excluding 0 because otherwise number will be 3-digited only).

Then hundred place can also be filled in 9 ways (including 0 and excluding the number placed in thousand place as repetition is not allowed).
Now ten’s place can be filled by any one of the remaining 8 digits in 8 ways

and then unit place can be filled in 7 ways.
∴ The required number of 4-digit numbers = 9 × 9 × 8 × 7 = 4536

Q.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:

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

For 3-digit even numbers, the unit’s place can be occupied by one of the 3 digits (2, 4 or 6) out of the given digits.

The digit cannot be repeated in 3-digit numbers and the unit place is occupied with a digit (2,4 or 6).

The unit place can be filled in ³P₁ = 3 ways by any digits from 2,4 or 6.

Hundreds, tens place can be filled by remaining any 5 digits.

Therefore, there will be as many 2-digit numbers as there are permutations of 5 different digits taken 2 at a time.
The remaining 5 digits can be arranged in the remaining 2 places in ⁵P₂ ways.
∴ By the multiplication rule, the required number of 3-digit even numbers = 3 × ⁵P₂ = 3 × 5 × 4 = 60.

Q.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:

4-digit numbers that can be formed using the digits 1, 2, 3, 4,5.

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,

For 4-digit numbers, we have to arrange the given 5 digits in 4 vacant places.

Therefore, the required number of 4-digit numbers = ⁵P₄ = 5 × 4 × 3 × 2 = 120 ways.

4-digit even numbers can be made using the digits 1, 2, 3, 4, 5 if no digit is repeated.

The unit place can be filled in 2 ways by any digits from 2 or 4.

The digit cannot be repeated in 4-digit numbers and the unit place is occupied with a digit(2 or 4) out of the given digits.

The digit cannot be repeated in 4-digit numbers and the unit place is occupied with a digit (2, or 4).

The unit place can be filled in ²P₁ = 2 ways by any digits from 2,or 4.

Thousands, hundreds, tens place can be filled by remaining any 4 digits.

Therefore, there will be as many 3-digit numbers as there are permutations of 4 different digits taken 3 at a time.

Therefore, the required number of 3-digit numbers = ⁴P₃

∴ By the multiplication rule,

The required number of 4-digit even numbers is 2 × ⁴P₃ = 2 × 4 × 3 × 2 = 48.

Q.5

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

Solution:

From a committee of 8 persons, chairman and a vice chairman are to be chosen assuming one person can not hold more than one position.

Out of 8 persons, a chairman can be chosen in 8 ways and then a vice chairman can be chosen in 7 ways.

∴ By the multiplication rule, the selection can be made in 8 × 7 = 56 ways

Second method :

Therefore,number of ways of choosing a chairman and a vice chairman is permutations of 8 different objects taken 2 at a time.

Therefore, we can arrange 8 persons in two vacant places in ⁸P₂ = 8 × 7 = 56 ways.

Q.6

Find n if ^n-1P₃ : ⁿP₃ = 1: 9.

Solution:

Q.7

Find r if

(i)  ⁵P_r = 2 ⁶P_r-1 

(ii) ⁵P_r = ⁶P_r-1

Solution:

Q.8

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

Solution:

There are 8 different letters in word EQUATION.

Therefore, words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once is permutations of 8 different letters taken 8 at a time.

The word EQUATION has 8 distinct letters which can be arranged among themselves in ⁸P₈ ways.
∴ The required number of words formed = ⁸P₈ = 8 ! = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 = 40320.

Q.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 the first letter is a vowel.

Solution:

The word MONDAY has 6 distinct letters.
(i)

4 letters out of 6 can be arranged in ⁶P₄ ways.
∴ The required number of words = ⁶P₄ = 6 × 5 × 4 × 3 = 360.
(ii)

6 letters can be arranged among themselves in ⁶P₆ ways.
∴ The required number of words = ⁶P₆ = 6 ! = 1 × 2 × 3 × 4 × 5 × 6 = 720.
(iii)

The first place (I) can be filled by anyone of the two vowels O or A in 2 ways.

The remaining 5 letters can be arranged in the remaining 5 places II to VI in ⁵P₅ = 5 ! ways.

∴ Words formed starting from vowel using 6 letters = 2 × 5 ! = 2 × 1 × 2 × 3 × 4 × 5 = 240

Q.10

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

Solution:

The word MISSISSIPPI has 11 letters, not all distinct.

In the given word MISSISSIPPI, I appears 4 times, S appears 4 times, M appears 1 time and P appear 2 times.

Now let us find the number of permutations in which the four I’s come together. Treating the four I’s as one letter (One object) for the time being.

(I I I I) S S S S P P M

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

These are 1 + 4 + 2 + 1 = 8 letters (Objects)

These 8 letters (objects) in which there are 4Ss and 2Ps can be arranged in

Hence, the number of permutations in which the four I’s do not come together =
= Total number of arrangements – Number of arrangements in which four I’s are together.
= 34650 – 840 = 33810.

Q.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 word PERMUTATIONS has 12 letters, not all distinct, T occurs twice.
(i)

For words starting with P and ending with S, we have to arrange the remaining 10 letters with T occuring twice in the 10 vacant places between them.

(ii)

The word PERMUTATIONS contains 5 distinct vowels E, U, A, I, O and 7 consonants in which T occurs twice.

Since the vowels have to occur together, we assume the group of vowels (EUAIO) as a single object.This single object together with 7 consonants become 8 objects, T occuring twice. These 8 objects can be arranged in
Corresponding to each of these arrangements, the 5 vowels, all distinct, can be arranged in ⁵P₅ = 5 !  ways.

Therefore, by multiplication principle, the required number of words

(iii)

Since there are always 4 letters between P and S, therefore, P and S can occupy either,

1st and 6th places

or 2nd and 7th places

or 3rd and 8th places

or 4th and 9th places

or 5th and 10th places

or 6th and 11th places

or 7th and 12th places.

Possible ways =7,

Also, P and S can be interchanged,

No. of permutations = 2 × 7 =14

The remaining 10 places can be filled with the remaining 10 letters, T occuring twice, in

= 7 × (1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 × 10)
= 7 × 720 × 56 × 90 = 25401600.

Exercise 7.4

Note: In problems on combinations (⇒ selections or groups) only number of things is important but not their order. So, we must have to apply combinations formulae in the following types of problems: (i) Straight lines and triangles formed by a given number of points (ii) Invitations (iii) Selections (iv) Groups (v) Committees (vi) Drawn.

Q.1

If ⁿC₈ = ⁿC₂, find ⁿC_2.

Solution:

Q.2

Determine n if
(i) ²ⁿC₃ : ⁿC₃ = 12 : 1
(ii) ²ⁿC₃  : ⁿC₃ = 11 : 1

Solution:

Q.3

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

Solution:

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

Given 21 points on a circle.

Now, a chord can be drawn by joining any two of the 21 given points.
∴ The required number of chords = ²¹C₂ 

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

Q.4

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

Solution:

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 4 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

Q.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:

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

Q.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:

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

∴ By multiplication principle, one ace and 4 other cards can be selected in ⁴C₁ × 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.

Q.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 4 bowlers?

Solution:

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.

Q.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:

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.

Q.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:

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.

Miscellaneous Exercise

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

2 vowels out of 3 can be selected in ³C₂ = 3 ways

Similarly,3 consonants out of 5 can be selected in ⁵C₃ = 10 ways.

∴ The number of choosing 2 vowels and 5 consonants would be ³C₂ × ⁵C₃ = 3 × 10 = 30

∴ The total number of ways of is 30.

Now, each of these 30 selections has 5 letters which can be arranged among themselves in ⁵P₅

= 5 ! = 1 × 2 × 3 × 4 × 5 = = 120 ways.
∴ The required number of different words is 30 × 120 = 3600.

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

Note: In the above question we first formed combinations (selections) and then permutations (arrangements) of those selections.

Q.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:

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

The word EQUATION has 8 distinct letters in which there are 5 vowels, namely E, U, A, I, O and 3 consonants, namely Q, T and N.

Since the vowels and consonants occur together, we assume the 5 vowels as one object (EUAIO) and the 3 consonants as another object (QTN).

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

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

Thus, these two objects can be arranged among themselves in 2 ! = 2 ways.
In each of these 2 arrangements, the 5 vowels can interchange places in 5 ! ways and the 3 consonants can interchange places in 3 ! ways.That is

The numbers of ways in which 5 vowels can be arranged are ⁵P₅ = 5 ! = 120 ways

Similarly, the numbers of ways in which 3 consonants can be arranged are ³P₃ = 3 ! = 6 ways
∴ By multiplication principle, the required number of words is 2 ! × 5 ! × 3 ! = 2 × 120 × 6 = 1440.

∴ The total number of ways in which vowel and consonant can appear together are 2 × ⁵P₅ × ³P₃

∴  = 2 × 120 × 6 = 1440

Q.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?

Solution:

(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)

Since at least 3 girls (i.e., 3 or more than 3) are to be there in every committee, therefore, the committee can consist of (a) 3 girls and 4 boys (b) 4 girls and 3 boys

3 girls and 4 boys can be selected in ⁴C₃ × ⁹C₄ ways.
4 girls and 3 boys can be selected in ⁴C₄ × ⁹C₃ ways.
∴ The required number of ways = ⁴C₃ × ⁹C₄ + ⁴C₄ × ⁹C₃

The total number of ways of making the committee are 504 + 84 = 588

(iii)

Since at most 3 girls (i.e., 3 or less than 3) are to be there in every committee, therefore, the committee can consist of
(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 ⁴C₃ × ⁹C₄ ways.
2 girls and 5 boys can be selected in ⁴C₂ × ⁹C₅ ways.
1 girl and 6 boys can be selected in ⁴C₁ × ⁹C₆ ways.
No girl and 7 boys can be selected in ⁴C₀ × ⁹C₇ ways.

∴ The required number of ways = ⁴C₃ × ⁹C₄ + ⁴C₂ × ⁹C₅ + ⁴C₁ × ⁹C₆ + ⁴C₀ × ⁹C₇

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

Q.4

If the different permutations of all the letters of the word EXAMINATION are listed as in a dictionary, how many words are there in this list before the first word starts with E?

Solution:

Total number of letters in this word is 11.

Writing the letters of the word EXAMINATION in alphabetic order, we have AAEIIMNNOTX

The required number of words before the first word starting with E is equal to the number of words which begin with A. Because only A appears before E in the above alphabetical order. When A is fixed in the first place, we have to arrange the remaining 10 letters in which there are two I’s and two N’s.

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

= words starting with letter A = 907200

Q.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:

Numbers divisible by 10 must have ‘0’ in the unit’s place.The remaining 5 digits can be arranged in the remaining 5 vacant places in ⁵P₅ = 5 ! ways.
∴ The required number of 6-digit numbers = 5 ! = 120.

Q.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:

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 = 10 ways.

Choosing 2 consonants out of 21 can be done in ²¹C₂ = 210 ways.

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

= ⁵C₂ × ²¹C₂ = 10 × 210 = 2100

Now, each of these 2100 selections has 4 letters which can be arranged among themselves in ⁴P₄ = 4 ! = 24 ways

Therefore, the required number of different words = 2100 × 24 = 50400

Q.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:

The various possibilities of selecting 8 questions are:

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

Or

5 questions from part 1 and 3 from part II

(∵ A student has to select at least 3 questions from each part.)

The required number of ways
= ⁵C₃ × ⁷C₅ + ⁵C₄ × ⁷C₄ + ⁵C₅ × ⁷C₃

Q.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:

We have a deck of 52 cards that has 4 kings.

The numbers of remaining cards are 48.

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 = = 4 × 194580 = 778320 ways

Q.9

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

Solution:

Given there is a total of 9 people.

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

4 women can sit in four places and ways they can be seated= ⁴P₄ = 24 ways

5 men can occupy 5 seats in 5 ways.

The number of ways in which these can be seated = ⁵P₅ = 120 ways

The total numbers of sitting arrangements possible are = 24 × 120 = 2880

Q.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:

According to given, the only two possibilities are:
(i) the particular 3 students join
(ii) the particular 3 students do not join.

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 = 170544

(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 = 646646

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

= 170544 + 646646 = 817190

Q.11

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

Solution:

The word ASSASSINATION has 13 letters, of which A appears 3 times, S appears 4 times, I appears 2 times, N appears 2 times and the rest are all different. Since all the
S’s are to occur together, we take them as a single object (SSSS). This single object together with 9 remaining letters become 10 objects (SSSS) AAA II NN TO which can be arranged in

The four S’s can be arranged among themselves in

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

The major concepts of Maths covered in Chapter 7 – Permutations and Combinations of NCERT Solutions for Class 11 include:

7.1 Introduction

7.2 Fundamental Principle of Counting

7.3 Permutations

7.3.1 Permutations when all the objects are distinct

7.3.2 Factorial notation

7.3.3 Derivation of the formula for ⁿP_r

7.3.4 Permutations when all the objects are not distinct objects

7.4 Combinations

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

The chapter Permutations and Combinations belongs to the unit Algebra, which adds up to 30 marks of the total 80 marks. There are 4 exercises along with a miscellaneous exercise in this chapter to help students understand the concepts related to Permutations and Combinations clearly. Some of the topics discussed in Chapter 7 of NCERT Solutions for Class 11 Maths are as follows:

1. The fundamental principle of counting is if an event can occur in m different ways, following which another event can occur in n different ways, then the total number of occurrences of the events in the given order is m × n.
2. The number of permutations of n different things taken r at a time, where repetition is not allowed, is denoted by ⁿP_r
3. n! = 1 × 2 × 3 × …×n
4. n! = n × (n – 1) !
5. The number of permutations of n different things, taken r at a time, where repetition is allowed, is n^r

Hence, we can conclude that studying the Permutations and Combinations of Class 11 enables the students to understand the Fundamental principle of counting, Factorial (n!), Permutations and combinations, derivation of Formulae for ⁿP_r and ⁿC_r and their connections and the simple applications of permutations and combinations.

Frequently Asked Questions (FAQs) on NCERT Solutions for Class 11 Maths Chapter 7

Q1

What are the crucial topics covered in the NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations?

The crucial topics covered in the NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations are
1. Introduction
2. Fundamental Principle of Counting
3. Permutations
4. Combinations

Q2

Discuss the fundamental concepts of Permutations and Combinations in the NCERT Solutions for Class 11 Maths Chapter 7.

Permutation is the several ways in which a number or a set of things can be arranged, whereas combination is a specific arrangement of different elements. Based on some of the primary concepts, the fundamental theorem of counting is the first one. For example, if an event is occurring in ‘n’ different ways and another event is happening in ‘m’ different ways, the net occurrences can be written in the order of n x m.
The second one explains that the number of permutations ‘n’ separate things at ‘r’ time, where the repetition does not take place, can be written as ⁿP_r and when repetition takes place is denoted as n^r.

Q3

Can I get the NCERT Solutions for Class 11 Maths Chapter 7 in PDF format?

Yes, students can download the NCERT Solutions for Class 11 Maths Chapter 7 in PDF format from ANAND CLASSESS website. The solutions for the exercise-wise problems are designed by the highly experienced faculty based on the latest CBSE Syllabus. Students can also cross-check their answers with the solutions PDF in order to get a clear idea about the other methods of solving complex problems effortlessly.

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