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 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

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 = ⁹P₃

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 ⁹P₃
= 9 x 504 = 4536.

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 ⁵P₂ ways.
∴ The required 3-digit even numbers
= 3 x ⁵P₂
= 60

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 ⁵P₄ways.
∴ The required 4 digit numbers = ⁵P₄ = 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 ⁴P₃ ways.
∴ The required 4-digit even numbers
= 2 x ⁴P₃ = 2 x 24 = 48

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

Question 6.
Find n if  ^n-1P₃: ⁿP₄ = 1 : 9.
Solution.

Question 7.
Find r if
(i) ⁵P_r = 2⁶P_r-1
(ii) ⁵P_r = ⁶P_r-1
Solution.

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
= ⁸P₈ = 8!
=40320

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
= ⁶P₄

(ii) When all letters are used at a time. Then the required number of words
= ⁶P₆ = 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 ⁵P₅ ways.
∴ The required number of words
= 2 x ⁵P₅
= 2 x 5! =240.

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

= 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 ways i.e. in 840 ways      … (ii)
Number of arrangements when four I’s do not come together = 34650 – 840 = 33810.

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 =

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

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