Permutations and Combinations Class 11 Commerce Maths 2 Chapter 6 Exercise 6.1 Answers Maharashtra Board
Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Permutations and Combinations Ex 6.1 Questions and Answers.
Std 11 Maths 2 Exercise 6.1 Solutions Commerce Maths
Question 1.
A teacher wants to select the class monitor in a class of 30 boys and 20 girls. In how many ways can he select a student if the monitor can be a boy or a girl?
Solution:
There are 30 boys and 20 girls in a class.
The teacher wants to select a class monitor from these boys and girls.
A boy can be selected in 30 ways and a girl can be selected in 20 ways.
∴ By using the fundamental principle of addition,
in a number of ways either a boy or a girl is selected as a class monitor = 30 + 20 = 50.
Question 2.
In question 1, in how many ways can the monitor be selected if the monitor must be a boy? What is the answer if the monitor must be a girl?
Solution:
(i) Since there are 30 boys in the class
∴ A boy monitor can be selected in 30 ways.
(ii) Since there are 20 girls in the class
∴ A girl monitor can be selected in 20 ways.
Question 3.
A Signal is generated from 2 flags by putting one flag above the other. If 4 flags of different colours are available, how many different signals can be generated?
Solution:
A signal is generated from 2 flags and there are 4 flags of different colours available.
∴ 1st flag can be any one of the available 4 flags.
∴ It can be selected in 4 ways.
Now, 2nd flag is to be selected for which 3 flags are available for a different signal.
∴ 2nd flag can be anyone from these 3 flags.
∴ It can be selected in 3 ways.
∴ By using the fundamental principle of multiplication,
Total number of ways in which a signal can be generated = 4 × 3 = 12
∴ 12 different signals can be generated.
Question 4.
How many two-letter words can be formed using letters from the word SPACE when repetition of letters
(i) is allowed
(ii) is not allowed
Solution:
A two-letter word is to be formed out of the letters of the word SPACE.
(i) When repetition of the letters is allowed
1st letter can be selected in 5 ways
2nd letter can be selected in 5 ways
∴ By using the fundamental principle of multiplication,
total number of 2-letter words = 5 × 5 = 25
(ii) When repetition of the letters is not allowed
1st letter can be selected in 5 ways
2nd letter can be selected in 4 ways
∴ By using the fundamental principle of multiplication,
total number of 2-letter words = 5 × 4 = 20
Question 5.
How many three-digit numbers can be formed from the digits 0, 1, 3, 5, 6 if repetitions of digits
(i) are allowed
(ii) are not allowed
Solution:
The three-digit number is to be formed from the digits 0, 1, 3, 5, 6
(i) When repetition of digits is allowed:
100’s place digit should be a non-zero number.
Hence, it can be anyone from digits 1, 3, 5, 6
∴ 100’s place digit can be selected in 4 ways.
0 can appear in 10’s and unit’s place and digits can be repeated.
∴ 10’s place digit can be selected in 5 ways and the unit’s place digit can be selected in 5 ways.
∴ By using the fundamental principle of multiplication,
the total number of three-digit numbers = 4 × 5 × 5 = 100
(ii) When repetition of digits is not allowed:
100’s place digit should be a non-zero number.
Hence, it can be anyone from digits 1, 3, 5, 6
∴ 100’s place digit can be selected in 4 ways
0 can appear in 10’s and unit’s place and digits can’t be repeated.
∴ 10’s place digit can be selected in 4 ways and the unit’s place digit can be selected in 3 ways
∴ By using the fundamental principle of multiplication,
total number of three-digit numbers = 4 × 4 × 3 = 48
Question 6.
How many three-digit numbers can be formed using the digits 2, 3, 4, 5, 6 if digits can be repeated?
Solution:
A 3-digit number is to be formed from the digits 2, 3, 4, 5, 6 where digits can be repeated.
∴ The unit’s place digit can be selected in 5 ways.
10’s place digit can be selected in 5 ways.
100’s place digit can be selected in 5 ways.
∴ By using fundamental principle of multiplication,
the total number of 3-digit numbers = 5 × 5 × 5 = 125
Question 7.
A letter lock has 3 rings and each ring has 5 letters. Determine the maximum number of trials that may be required to open the lock.
Solution:
A letter lock has 3 rings, each ring containing 5 different letters.
∴ A letter from each ring can be selected in 5 ways.
∴ By using fundamental principle of multiplication,
the total number of trials that can be made = 5 × 5 × 5 = 125
Out of these 124 wrong attempts are made and in the 125th attempt,
the lock gets opened, for a maximum number of trials.
∴ A maximum number of trials required to open the lock is 125.
Question 8.
In a test that has 5 true/false questions, no student has got all correct answers and no sequence of answers is repeated. What is the maximum number of students for this to be possible?
Solution:
For a set of 5 true/false questions, each question can be answered in 2 ways.
∴ By using fundamental principle of multiplication,
the total number of possible sequences of answers = 2 × 2 × 2 × 2 × 2 = 32
Since no student has written all the correct answers.
∴ Total number of sequences of answers given by the students in the class = 32 – 1 = 31
Also, no student has given the same sequence of answers.
∴ Maximum number of students in the class = Number of sequences of answers given by the students = 31
Question 9.
How many numbers between 100 and 1000 have 4 in the unit’s place?
Solution:
Numbers between 100 and 1000 are 3-digit numbers.
A 3-digit number is to be formed from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 where the unit place digit is 4.
Since Unit’s place digit is 4.
∴ it can be selected in 1 way only.
10’s place digit can be selected in 10 ways.
For 3-digit number 100’s place digit should be a non-zero number.
∴ 100’s place digit can be selected in 9 ways.
∴ By using fundamental principle of multiplication,
total number of numbers between 100 and 1000 which have 4 in the units place = 1 × 10 × 9 = 90
Question 10.
How many numbers between 100 and 1000 have the digit 7 exactly once?
Solution:
Numbers between 100 and 1000 are 3-digit numbers.
A 3-digit number is to be formed from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, where exactly one of the digits is 7.
When 7 is in the unit’s place
The unit’s place digit is 7.
∴ it can be selected in 1 way only.
10’s place digit can be selected in 9 ways.
100’s place digit can be selected in 8 ways.
∴ total number of numbers which have 7 in the unit’s place = 1 × 9 × 8 = 72
When 7 is in 10’s place
The unit’s place digit can be selected in 9 ways.
10’s place digit is 7
∴ it can be selected in 1 way only.
100’s place digit can be selected in 8 ways.
∴ total number of numbers which have 7 in 10’s place = 9 × 1 × 8 = 72
When 7 is in 100’s place
The unit’s place digit can be selected in 9 ways.
10’s place digit can be selected in 9 ways.
100’s place digit is 7
∴ it can be selected in 1 way.
∴ total numbers which have 7 in 100’s place = 9 × 9 × 1 = 81
∴ total number of numbers between 100 and 1000 having digit 7 exactly once = 72 + 72 + 81 = 225.
Question 11.
How many four-digit numbers will not exceed 7432 if they are formed using the digits 2, 3, 4, 7 without repetition?
Solution:
Among many set’s of digits, the greatest number is possible when digits are arranged in descending order.
∴ 7432 is the greatest number, formed from the digits 2, 3, 4, 7.
∴ Since a 4-digit number is to be formed from the digits 2, 3, 4, 7, where repetition of the digit is not allowed.
∴ 1000’s place digit can be selected in 4 ways.
100’s place digit can be selected in 3 ways.
10’s place digit can be selected in 2 ways.
The unit’s place digit can be selected in 1 way.
∴ Total number of numbers not exceeding 7432 that can be formed from the digits 2, 3, 4, 7
= Total number of four-digit numbers formed from the digits 2, 3, 4, 7
= 4 × 3 × 2 × 1
= 24
Question 12.
If numbers are formed using digits 2, 3, 4, 5, 6 without repetition, how many of them will exceed 400?
Solution:
Case I: Three-digit numbers with 4 occurring in hundred’s place:
100’s place digit can be selected in 1 way.
Ten’s place can be filled by any one of the numbers 2, 3, 5, 6.
∴ 10’s place digit can be selected in 4 ways.
The unit’s place digit can be selected in 3 ways.
∴ total number of numbers which have 4 in 100’s place = 1 × 4 × 3 = 12
Case II: Three-digit numbers more than 500
100’s place digit can be selected in 2 ways.
10’s place digit can be selected in 4 ways.
Unit’s place digit can be selected in 3 ways.
∴ total number of three digit numbers more than 500 = 2 × 4 × 3 = 24
Case III: Number of four digit numbers formed from 2, 3, 4, 5, 6
Since, repetition of digits is not allowed
∴ total four digit numbers formed = 5 × 4 × 3 × 2 = 120
Case IV: Number of five digit numbers formed from 2, 3, 4, 5, 6
Since, repetition of digits is not allowed
∴ total five digit numbers formed = 5 × 4 × 3 × 2 × 1 = 120
∴ total number of numbers that exceed 400 = 12 + 24 + 120 + 120 = 276
Question 13.
How many numbers formed with the digits 0, 1, 2, 5, 7, 8 will fall between 13 and 1000 if digits can be repeated?
Solution:
Case I: 2-digit numbers more than 13, less than 20, formed from the digits 0, 1, 2, 5, 7, 8
Number of such numbers = 3
Case II: 2-digit numbers more than 20 formed from 0, 1, 2, 5, 7, 8
Ten’s place digit is selected from 2, 5, 7, 8.
∴ Ten’s place digit can be selected in 4 ways.
Unit’s place digit is anyone from 0, 1, 2, 5, 7, 8
∴ The unit’s place digit can be selected in 6 ways.
Using the multiplication principle,
the number of such numbers (repetition allowed) = 4 × 6 = 24
Case III: 3-digit numbers formed from 0, 1, 2, 5, 7, 8
100’s place digit is anyone from 1, 2, 5, 7, 8.
∴ 100’s place digit can be selected in 5 ways.
As digits can be repeated, the 10’s place and unit’s place digits are selected from 0, 1, 2, 5, 7, 8
∴ 10’s place and unit’s place digits can be selected in 6 ways each.
Using multiplication principle,
the number of such numbers (repetition allowed) = 5 × 6 × 6 = 180
All cases are mutually exclusive and exhaustive.
∴ Required number = 3 + 24 + 180 = 207
Question 14.
A school has three gates and four staircases from the first floor to the second floor. How many ways does a student have to go from outside the school to his classroom on the second floor?
Solution:
A student can go inside the school from outside in 3 ways and from the first floor to the second floor in 4 ways.
∴ Number of ways to choose gates = 3
Number of ways to choose staircase = 4
∴ By using fundamental principle of multiplication,
number of ways in which a student has to go from outside the school to his classroom = 4 × 3 = 12
Question 15.
How many five-digit numbers formed using the digit 0, 1, 2, 3, 4, 5 are divisible by 3 if digits are not repeated?
Solution:
For a number to be divisible by 3.
The sum of digits must be divisible by 3.
Given 6 digits are 0, 1,2, 3, 4, 5.
Sum of 1, 2, 3, 4, 5 = 15, which is divisible by 3.
∴ There are two cases of 5 digit numbers formed from 0, 1, 2, 3, 4, 5 and divisible by 3.
Either 3 is selected in 5 digits (and 0 not selected) or 3 is not selected in 5 digits (and 0 is selected)
Case I:
3 is not selected (and 0 is selected) i.e., the digits are 0, 1, 2, 4, 5.
10000’s place digit can be selected in 4 ways (as 0 cannot appear).
As digits are not repeated, 1000’s place digit can be selected in 4 ways.
100’s place digit can be selected in 3 ways.
10’s place digit can be selected in 2 ways.
The unit’s place digit can be selected in 1 way.
∴ Using multiplication theorem,
Number of 5-digit number formed from 0, 1, 2, 4, 5 (with no repetition of digits) = 4 × 4 × 3 × 2 × 1 = 96
Case II:
3 is selected (and 0 is not selected) i.e., 1, 2, 3, 4, 5
10000’s place digit can be selected in 5 ways.
1000’s place digit can be selected in 4 ways.
100’s place digit can be selected in 3 ways.
10’s place digit can be selected in 2 ways.
The unit’s place digit can be selected in 1 way.
Using multiplication theorem,
Number of 5-digit numbers formed from 1, 2, 3, 4, 5 = 5 × 4 × 3 × 2 × 1 = 120
Both the cases are mutually exclusive and exhaustive.
∴ Required number = 96 + 120 = 216