NCERT Solutions for Class 11 Maths chapter-7 Permutation And Combination Exercise 7.3
NCERT Solutions for Class-11 Maths Chapter-7 Permutation and Combination
NCERT Solutions For Class 11 Maths Chapter-7 Permutation and Combination prepared by the expert of Physics Wallah score more with Physics Wallah NCERT Class 11 maths solutions. You can download solution of all chapters from Physics Wallah NCERT solutions of class 11.
NCERT Solutions for Class-11 Maths Exercise 7.3
Question 1. How many 3-digit numbers can be formed by using the digits 1 to 9 if no digits is repeated?
Solution :
Total number of digits = 9 and Number of digits used without repetition = 3
Question 2. How many 4-digit numbers are there with no digit repeated?
Solution :
Total number of digits = 10 and Number of digits used without repetition = 4
0 cannot be filled in the fourth place therefore number of permutations for fourth place = 9
Hence total number of permutations = 9 × 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 :
Total number of digits = 6 and unit place can be filled with any one of the digits 2, 4, 6.
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 :
Total number of digits = 5 and Number of digits used without repetition = 4
Now, unit place can be filled with any one of the digits 2, 4 for even number.
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 :
Total number of persons = 8 and Number of persons used without repetition = 2
Question 6. Find n if n−1 P 3 : n P 4 = 1:9
Solution :
Given:
n−1
P
3
:
n
P
4
= 1:9
Question 7. Find r if
(i) 5 P r = 2 6 P r−1
(ii) 5 P r = 6 P r−1
Solution :
(i) Given:
5
P
r
= 2
6
P
r−1
(ii) Given: 5 P r = 6 P r−1
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 different letters in the given word is 8.
Thus, the number of words than can be formed without repetition is number of permutations of 8 different objects taken 8 at a time =
8
P
8
=8!
Therefore, number of words formed = 8! = 40320.
Question 9. How many words, with or without meaning can be formed using all 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 by first letter is a vowel?
Solution :
Total number of letters in word MONDAY = 6 and number of vowels = 2
(i) Number of letters used = 4
(ii) Number of letters used = 6
Number of permutations = 6!=6×5×4×3×2×1=720= 720
(iii) First letter is vowel.
There are two different vowels in the word MONDAY which occupies the rightmost place of the words formed. Hence, there are 2 ways.
Since, it is without repetition and the rightmost place is occupied, the remaining five vacant places can be filled by 5 different letters. Hence, 5! Ways.
Therefore, number of words that can be formed = 5!×2=120×2=240
Question 10. In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?
Solution :
I appears 4 times, S appears 4 times, P appears 2 times and M appears one time in the given word
Thus, number of permutations of the given word
Since, there are 4 I’s in the given word, they can be treated as a single object. This single object with the remaining 7 objects will together be 8 objects.
These 8 objects in which 4 S’s and 2 P’s is arranged in
= 840 ways
Thus, number of ways where I’s occur together = 840
Therefore, number of permutations where I’s do not occur together is =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 :
Total numbers in the word PERMUTATIONS = 12 and here T = 2
(i) First letter is P and last letter is S which are fixed.
Therefore, remaining 10 letters are to be arranged between P and S.
⇒ Number of permutations =
= 1814400
(ii) There are 5 vowels in the word PERMUTATIONS. All vowels can be put together.
⇒ Number of permutations of all vowels together = 120
Now consider the 5 vowels together as one letter. SO the number of letters in the word when all vowels are together = 8
∴ Number of permutations =
(iii) P and S are on 1st and 6 th places P and S are on 2nd and 7th places
P and S are on 3rd and 8 th places P and S are on 4th and 9th places
P and S are on 5th and 10 th places P and S are on 6th and 11th places
P and S are on 7th and 12 th places
Now, P and S can be put in 7 ways and also P and S can interchange their positions.
∵ Number of permutations = 2 × 7= 14
Now the remaining 10 places can be filled with remaining 10 letters.
∴ Number of permutations =
Therefore, total number of permutations = 14 × 1814400 = 25401600
