Chapter 7 Permutations and Combinations Ex 7.3 Solutions
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-1P3: nP3 =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
⇒ 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