## Permutations and Combinations NCERT Solutions Class 11 Maths Chapter 7 Exercise 7.1 7.2 7.3 7.4 Miscellaneous Free pdf Notes Study Material download-Anand Classes |

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 2^{n}.

# 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

# Compute**Solution:**

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

(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 = ^{9}P_{3} = 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 ^{10}P_{4}.

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 ^{9}P_{3} ways.

∴ The required number of 4-digit numbers = ^{10}P_{4} – ^{9}P_{3}

= 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 ^{3}P_{1} = 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 ^{5}P_{2} ways.

∴ By the multiplication rule, the required number of 3-digit even numbers = 3 × ^{5}P_{2} = 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 = ^{5}P_{4} = 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 ^{2}P_{1} = 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 = ^{4}P_{3}

∴ By the multiplication rule,

The required number of 4-digit even numbers is 2 × ^{4}P_{3} = 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 ^{8}P_{2} = 8 × 7 = 56 ways.

**Q.6**

**Find n if **^{n-1}P_{3 }: ^{n}P_{3} = 1: 9.

^{n-1}P

_{3 }:

^{n}P

_{3}= 1: 9.

**Solution:**

**Q.7**

**Find r if**

**(i) **^{5}P_{r} = 2 ^{6}P_{r-1}

^{5}P

_{r}= 2

^{6}P

_{r-1}

**(ii) **^{5}P_{r} = ^{6}P_{r-1}

^{5}P

_{r}=

^{6}P

_{r-1}

**Solution:**