Class 12 Permutation and Combination Exercise 1.1 Solutions | Basic Mathematics Sukunda Pustak Bhawan
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A football stadium has four entrance gates and nine exits. In how many different ways can a man enter and leave the stadium?
Atith Adhikari
·
2 years ago
Solution
A man can enter the stadium from any one of the four entrance gates (m = 4) and can leave the stadium from any one of the nine exit gates (n = 9).
By the basic principle of counting, a man can enter and leave the stadium in $\rm m \times n = 4x9 = 36$ different ways.
There are six doors in a hostel. In how many ways can a student enter the hostel and leave by a different door?
Atith Adhikari
·
2 years ago
Solution
A student can enter the hostel by any one of the six doors (m = 6).
To exit the hostel from a different door, he cannot exit from the same door he entered. So, the total choices for exit doors are five (n = 5).
By the basic principle of counting, a student can enter and leave the hostel by a different door in $\rm m \times n = 6x5 = 30$ different ways.
In how many ways can a man send three of his children to seven different colleges of a certain town?
Atith Adhikari
·
2 years ago
Solution
A man has seven college choices for his first child (m = 7).
To send his second child to a different college, he cannot send him to the same college where he sent his first child. So, the second child can go into any one of the six remaining colleges (n = 6).
Similarly, the third child can choose from the remaining five colleges (o = 5).
By the basic principle of counting, a man can send three of his children to seven different colleges in a certain town in $\rm m \times n \times o = 7x6x5 = 210$ ways.
Suppose there are five main roads between the cities A and B. In how many ways can a man go from a city to the other and return by a different road?
Atith Adhikari
·
2 years ago
Solution
A man can from city A to city B via any one of the five main roads (m = 5).
During his return, if he wishes to come back through a different road than the previous, then he has to choose from the remaining four main roads (n = 4).
By the basic principle of counting, a man can go from one city to the other and return by a different road in $\rm m \times n = 5 \times 4 = 20$ ways.
There are five main roads between the cities A and B and 4 between B and C. In how many ways can a person drive from A to C and return without driving on the same road twice?
Atith Adhikari
·
2 years ago
Solution
Let us consider the man goes from A-B-C and returns from C-B-A.
When he goes from A to B, he can choose one of the five main roads (m = 5). When going from B to C, he can choose one of the four main roads (n = 4).
During his return, he will have one less road available if he does not drive on the same road twice. So, from C to B, he has three choices (o = 3). Similarly, from B to A, he has four choices (p = 4).
By the basic principle of counting, a man can drive from A to C and return without driving on the same road twice in $\rm m \times n \times o \times p = 5 \times 4 \times 3 \times 4 = 240$ ways.
How many numbers of three digits less than 500 can be formed from the integers $\rm 1, 2, 3, 4 , 5, 6\; $?
Atith Adhikari
·
2 years ago
Solution
A three-digit number less than 500 is anywhere between 99 and 500.
The hundreds place can only be occupied by four of the six integers (1, 2, 3, 4) to make it less than 500 (m = 4).
The tenth place can be filled by any of the remaining five integers (n = 5).
Similarly, the one's place can be filled by four integers (o = 4).
By the basic principle of counting, the number of three digits less than 500 that can be formed from the integer 1,2,3,4,5,6 is $\rm m \times n \times o = 4x5x4 = 80$.
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