NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.3 are part of NCERT Solutions for Class 11 Maths. Here we have given NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.3.

- Permutations and Combinations Class 11 Ex 7.1
- Permutations and Combinations Class 11 Ex 7.2
- Permutations and Combinations Class 11 Ex 7.4

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 11 |

Subject |
Maths |

Chapter |
Chapter 7 |

Chapter Name |
Permutations and Combinations |

Exercise |
Ex 7.3 |

Number of Questions Solved |
11 |

Category |
NCERT Solutions |

## 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 digits are 9. We have to form 3 digit numbers without repetition.

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

\(=\frac { 9! }{ 6! } =9\times 8\times 7=504\)

**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 digits 0 to 9. In four digit numbers 0 is not taken at thousand’s place, so thousand’s place can be filled in 9 different ways. After filling thousand’s place, 9 digits are left. The remaining three places can be filled in 9P3 ways.

So the required 4-digit numbers

= 9 x ^{9}P_{3}

= 9 x 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 3-digit even numbers

= 3 x ^{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 from digits 1 to 5 in ^{5}P_{4}ways.

∴ The required 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 4-digit even numbers

= 2 x ^{4}P_{3} = 2 x 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

**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.**

No. of letters in the word EQUATION = 8

∴ No. 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.**

No. of letters in the word MONDAY = 6

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

Then, the required number of words

= ^{6}P_{4}

\(=\frac { 6! }{ 2! } =6\times 5\times 4\times 3=360\)

**(ii)** When all letters are used at a time. Then the required number of words

= ^{6}P_{6} = 6!

= 720

**(iii)** All letters are used but first letter is a vowel.

So the first letter can be either A or O.

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

∴ The required number of words

= 2 x ^{5}P_{5}

= 2 x 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, of which I appears 4 times, S appears 4 times, P appears 2 times & M appears 1 time.

∴ The required number of arrangements

\(=\frac { 11! }{ 4!4!2! } =\frac { 11\times 10\times 9\times 8\times 7\times 6\times 5\times 4! }{ 4\times 3\times 2\times 2\times 4! } \)

= 10 x 10 x 9 x 7 x 5 = 34650 … **(i)**

When four I’s come together, we treat them as a single object. This single object with 7 remaining objects will account for 8 objects. These 8 objects in which there are 4S’s & 2P’s

can be rearranged in \(\frac { 8! }{ 4!2! } \) ways i.e. in 840 ways … **(ii)**

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 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 words = \(\frac { 10! }{ 2! } \)

\(=\frac { 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2! }{ 2! } =1814400\)

**(ii)** When vowels are taken together i.e E U A I O we treat them as a single object. This single object with remaining 7 objects will account for 8 objects, in which there w are 2Ts, which can be rearranged in \(\frac { 8! }{ 2! } =20160\) ways. Corresponding to each of these arrangements the 5 vowels E, U, A, I, O can be rearranged in 5! = 120 ways. Therefore, by multiplication principle, the required number of arrangements = 20160 x 120 = 2419200.

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

∴ P & S can be at

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

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

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

4^{th} & 9^{th} place

5^{th} & 10^{th} place

6^{th} & 11^{th} place

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

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

The remaining 10 letters with 2T’s, can be arranged in \(\frac { 10! }{ 2! } =1814400\) ways.

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

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