**NCERT Solutions for Class 11 Maths Chapter 7 Permutations and
Combinations Ex 7.3**

NCERT Solutions for Class
11 Maths Chapter 7
Permutations and Combinations Ex 7.3 are the part of NCERT Solutions for Class
11 Maths. Here you can find the NCERT Solutions for Class 11 Maths chapter 7
Permutations and Combinations Ex 7.3.

**Ex 7.3
Class 11 Maths Question 1.**

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

**Solution.**

Total number of digits are 9. We have to form 3-digit numbers without
repetition.

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

= 9!/6! = 9 × 8 × 7 = 504

**Ex 7.3 Class 11
Maths Question 2.**

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

**Solution.**

The 4-digit numbers are formed from the digits 0 to 9. In 4-digit numbers 0 is
not taken at the thousands place, so the thousands place can be filled in 9
different ways. After filling thousands place, 9 digits are left. The remaining
three places can be filled in ^{9}P_{3} ways.

So, the required number of 4-digit numbers

= 9 × ^{9}P_{3}

= 9 × 504 =
4536

**Ex 7.3 Class 11
Maths Question 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.**

For 3-digit even numbers, unit place can be filled by 2, 4, 6, i.e., in 3 ways.
Then, the remaining two places can be filled in ^{5}P_{2} ways.

∴ The required number of 3-digit even
numbers

= 3 × ^{5}P_{2}

= 60

**Ex 7.3 Class 11 Maths Question
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 4-digit
numbers can be formed using the digits 1, 2, 3, 4, 5 in ^{5}P_{4
}ways.

∴ The required number of 4-digit
numbers = ^{5}P_{4} = 120

For 4-digit even numbers unit place can be filled by 2, 4,
i.e., in 2 ways. Then, the remaining three places can be filled in ^{4}P_{3} ways.

∴ The required number of 4-digit even
numbers

= 2 × ^{4}P_{3} =
2 × 24 = 48

**Ex 7.3 Class 11
Maths Question 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, we can choose a chairman and a vice chairman in ^{8}P_{2 }ways.

Thus, the
number of ways we can choose a chairman and vice chairman assuming one person
cannot hold more than one position = ^{8}P_{2 }= 56.

**Ex 7.3 Class 11
Maths Question 6.**

Find n if ^{n-1}P

_{3 }:

^{n}P

_{4}= 1 : 9.

**Solution.**

**Ex 7.3 Class 11
Maths Question 7.**

Find r if**(i)**

^{5}P

_{r}= 2

^{6}P

_{r-1}

**(ii)**

^{5}P

_{r}=

^{6}P

_{r-1}

**Solution.**

**Ex 7.3 Class 11
Maths Question 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.**

The number of letters in the word EQUATION = 8

∴ The number of words that can be formed

= ^{8}P_{8} = 8!

= 40320

**Ex 7.3 Class 11 Maths Question 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.**

The number of letters in the word MONDAY = 6

**(i)** When
4 letters are used at a time.

Then, the required number of words made = ^{6}P_{4}

= 6!/2! = 6 × 5 × 4 × 3 = 360

**(ii)** When all the letters are used at a
time.

Then, the required number of
words make = ^{6}P_{6} = 6!

= 720

**(iii)** When all the letters are used but
first letter is a vowel.

So, the first letter can be either A or O.

Thus, there are 2 ways to fill the first letter and remaining places can be
filled in ^{5}P_{5} ways.

∴ The required number of words made = 2 × ^{5}P_{5}

= 2 × 5! = 240

**Ex 7.3 Class 11 Maths Question 10.**

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

**Solution.**

There are 11 letters in the word MISSISSIPPI, of which I appears 4 times, S
appears 4 times, P appears 2 times and M appears 1 time.

∴ The required number of arrangements

= 11!/(4!4!2!) = (11 × 10 × 9 × 8 × 7 × 6 × 5)/(4 × 3 × 2 × 1 × 2 × 1)

= 34650
… **(i)**

When four I’s come together, we treat them as a single letter. So, the total
number of letters in the word will be 8. These 8 letters in which there are four
S’s and two P’s

can be rearranged in 8!/4!2! ways, i.e., in 840 ways.
… **(ii)**

Hence, the number of arrangements when four I’s do not come together

= 34650 – 840 = 33810

**Ex 7.3 Class 11 Maths Question 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.**

There are 12 letters in the word PERMUTATIONS of which T appears 2 times.

**(i)** When
words start with P and end with S, then there are 10 letters to be arranged of
which T appears 2 times.

∴ The required number of words = 10!/2!

= 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 = 1814400

**(ii)** When vowels are taken together, i.e., E U A I O
we treat them as a single letter. This single letter with remaining 7 letters
will account for 8 letters, in which there are two Ts, which can be rearranged
in 8!/2! = 20160 ways.

Corresponding to each of these arrangements, the 5
vowels E, U, A, I, O can be rearranged in 5! ways = 120 ways. Therefore, by
multiplication principle, the required number of arrangements = 20160 × 120 =
2419200.

**(iii)** When there are always 4 letters between P and S

∴ P and S can be at

1^{st} and 6^{th} place

2^{nd} and 7^{th} place

3^{rd }and 8^{th} place

4^{th} and 9^{th} place

5^{th} and 10^{th} place

6^{th} and 11^{th} place

7^{th} and 12^{th} place.

So, P and S will be placed in 7 ways and can be arranged in 7 × 2! ways = 14
ways

The remaining 10 letters with two T’s, can be arranged in 10!/2! ways = 1814400 ways.

∴ The required number of arrangements = 14
× 1814400 = 25401600.

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