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Chapter 7 Permutations and Combinations Ex-7.3 Interview Questions Answers

Question 1 : How many 3-digit numbers can be formed byusing the digits 1 to 9 if no digit is repeated?

Answer 1 :


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

Answer 2 :

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.

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?

Answer 3 :

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.

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?

Answer 4 :

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.

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?

Answer 5 :

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

Question 6 : Find n if n-1P3nP3 =1: 9.

Answer 6 :


Question 7 :

Find r if

(i)5Pr = 26Pr-1 

(ii) 5Pr = 6Pr-1

Answer 7 :


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?

Answer 8 :

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

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?

Answer 9 :

(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.

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

Answer 10 :

Total number ofletters in MISSISSIPPI =11

Letter Number ofoccurrence

M

1

I

4

S

4

P

2

 Number ofpermutations =

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


Selected

 

Chapter 7 Permutations and Combinations Ex-7.3 Contributors

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