Q1Written answer
How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?
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NCERT · Class 11 · Maths
Chapter 7: Permutations and Combinations Exercise 7.3
Practice each question, then open the answer or worked solution.
Questions
11 questionsQ2Written answer
How many 4-digit numbers are there with no digit repeated?
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Worked solution
To find the four-digit number (digits do 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
The 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.
Q3Written answer
How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?
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Worked solution
An even number means that the last digit should be even.
Number of possible digits at one’s place = 3 (2, 4 and 6)
⇒ Number of permutations=
One of the digits is taken at one’s place; the number of possible digits available = 5
⇒ Number of permutations=
Therefore, the total number of permutations =3 × 20=60
Q4Written answer
Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, and 5 if no digit is repeated. How many of these will be even?
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Worked 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 in one’s place.
The possibility of (2, 4) at one’s place = 2/5 = 0.4
The total number of even numbers = 120 × 0.4 = 48
Q5Written answer
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?
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Worked solution
Total number of people in committee = 8
Number of positions to be filled = 2
⇒ Number of permutations =
Q6Written answer
Find n if n-1P3: nP3 = 1: 9.
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Q7Written answer
Find r if
(i)5Pr = 26Pr-1
(ii) 5Pr = 6Pr-1
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Q8Written answer
How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?
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Worked solution
Total number of different letters in EQUATION = 8
Number of letters to be used to form a word = 8
⇒ Number of permutations =
Q9Written answer
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 the first letter is a vowel
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Worked 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, the remaining places = 5
Remaining letters to be used =5
⇒ Number of permutations =
Therefore, the total number of permutations = 2 × 120 =240
Q10Written answer
In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?
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Worked solution
Total number of letters in MISSISSIPPI = 11
Letter number of occurrence
| M | 1 |
| I | 4 |
| S | 4 |
| P | 2 |
⇒ 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, the total number of permutations where four I’s don’t come together = 34650-840=33810