NCERT Solutions Class 11th Maths Chapter – 7 Permutations and Combinations Exercise 7.3

NCERT Solutions Class 11th Maths Chapter – 7 Permutations and Combinations Exercise 7.3

NCERT Solutions Class 11th Maths Chapter – 7 Permutations and Combinations Exercise 7.3

?Chapter – 7?

✍Permutations and Combinations✍

?Exercise 7.3?

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

Solution:

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

Solution:
To find four digit number (digits does not repeat)
Now we will have 4 places where 4 digits are to be put.
So, at thousand’s place = There are 9 ways as 0 cannot be at thousand’s place = 9 ways
At hundredth’s place = There are 9 digits to be filled as 1 digit is already taken = 9 ways
At ten’s place = There are now 8 digits to be filled as 2 digits are already taken = 8 ways
At unit’s place = There are 7 digits that can be filled = 7 ways
Total Number of ways to fill the four places = 9 × 9 × 8 × 7 = 4536 ways.
So a total of 4536 four digit numbers can be there with no digits repeated.

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:
Even number means that last digit should be even,
Number of possible digits at one’s place = 3 (2, 4 and 6)
⇒ Number of permutations=

One of digit is taken at one’s place, Number of possible digits available = 5
⇒ Number of permutations=

Therefore, total number of permutations =3 × 20=60.

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 possible for choosing = 5
Number of places for which a digit has to be taken = 4
As there is no repetition allowed,
⇒ Number of permutations =

The number will be even when 2 and 4 are at one’s place.
The possibility of (2, 4) at one’s place = 2/5 = 0.4
Total number of even number = 120 × 0.4 = 48.

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 people in committee = 8
Number of positions to be filled = 2
⇒ Number of permutations =

6. Find n if ^n-1P₃: ⁿP₃ = 1: 9.

Solution:

7. Find r if

(i)⁵P_r = 2⁶P_r-1 

(ii) ⁵P_r = ⁶P_r-1

Solution:

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:
Total number of different letters in EQUATION = 8
Number of letters to be used to form a word = 8
⇒ Number of permutations =

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:
(i) Number of letters to be used =4
⇒ Number of permutations =

(ii) Number of letters to be used = 6
⇒ Number of permutations =

(iii) Number of vowels in MONDAY = 2 (O and A)
⇒ Number of permutations in vowel =

Now, remaining places = 5
Remaining letters to be used =5
⇒ Number of permutations =

Therefore, total number of permutations = 2 × 120 =240.

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

Solution:
Total number of letters in MISSISSIPPI =11
Letter Number of occurrence

⇒ Number of permutations =

We take that 4 I’s come together, and they are treated as 1 letter,
∴ Total number of letters=11 – 4 + 1 = 8
⇒ Number of permutations =

Therefore, total number of permutations where four I’s don’t come together = 34650-840=33810

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:
(i) Total number of letters in PERMUTATIONS =12
Only repeated letter is T; 2times
First and last letter of the word are fixed as P and S respectively.
Number of letters remaining =12 – 2 = 10
⇒ Number of permutations =

(ii) Number of vowels in PERMUTATIONS = 5 (E, U, A, I, O)
Now, we consider all the vowels together as one.
Number of permutations of vowels = 120
Now total number of letters = 12 – 5 + 1= 8
⇒ Number of permutations =

Therefore, total number of permutations = 120 × 20160 = 2419200

(iii) Number of places are as 1 2 3 4 5 6 7 8 9 10 11 12
There should always be 4 letters between P and S.
Possible places of P and S are 1 and 6, 2and 7, 3 and 8, 4 and 9, 5 and 10, 6 and 11, 7 and 12
Possible ways =7,
Also, P and S can be interchanged,
No. of permutations =2 × 7 =14
Remaining 10 places can be filled with 10 remaining letters,
∴ No. of permutations =

Therefore, total number of permutations = 14 × 1814400 =25401600.