UNIT 1: ALGEBRA

Permutation and Combination Class 12 Basic Maths 




Exercise 1.1

1. A football stadium has four entrance gates and nine exists. In how many different ways can a man enter and leave the stadium? 
 Soln: 
The man can enter the stadium in 4 different ways and can leave the stadium in 9 different ways.
 Now, bybasic principle of counting, he can enter and leave the stadium in 9 × 4 = 36 ways.  

2. There are six doors in a hostel. In how many ways can a student enter the hostel and leave by a different door?
Soln:
The student can enter in the hostel in 6 different ways and leave in 5 different ways. 
 Now, by the basic principle of counting, he can enter and leave by 6×5 = 30 ways. 

3. In how many ways can a man send three of his children to seven different colleges of a certain town?
Sonn:
The man can send one of his children to 7 different colleges, next to 6 colleges and next to 5 colleges. Now, by the basic principle of counting, three of his children can send to seven different colleges of a certain town by 7 × 6 × 5 = 210 ways

4. 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?
Soln:
The man has 5 option while going from a city to other and 4 while returning. Now, by the basic principle of counting, the no of ways that a man can go from a city to other and return by a different is 
5 × 4 = 20  

5. There are five main roads between the cities A and B and 4 between B and C. In how many ways can person drive from A to C and return without driving on the same road twice? 
 Soln:
The man has 5 option while going from city A to B and 4 option while going from B to C. 
 Similarly, man has 3 option while returning from C to B and 4 option while returning from B to A. Now, by the basic principle of counting, total no. of ways of driving on the road = 5× 4 × 3 ×4 =240

6. How many numbers of at least three different digits can be formed from the integers 1, 2, 3, 4, 5, 6,?
Soln: The numbers are of the three digits, four digits, five digits and six digits. Now, by the basic principle of counting,
 three-digit numbers can be formed in 6 × 5 × 4 = 120 ways, 
 four-digit numbers can be formed in 6 × 5 × 4 × 3 = 360 ways, 
five-digit numbers can be formed in 6× 5 × 4 × 3 × 2 = 720 ways.
 six-digit numbers can be formed in 6 × 5 × 4×3×2×1 = 720 ways. 
  Total no. of ways for selecting the numbers of at least 3 digits = 120 + 360 + 720 + 720 = 1920.

7. How many numbers of three digits less than 500 can be formed form the integers 1, 2, 3, 4, 5, 6? Soln: 
The required three digit numbers less than 500 should be started from 1 or 2 or 3 or 4. 
 So, hundreds place have 4 choices. 
Then tens and unit place have 5 and 4 choices 
 Total no. of ways of forming the numbers less than 500 = 4 × 5 × 4 = 80. 

 8. Of the numbers formed by using all the figures 1, 2, 3, 4, 5 only once, how many are even? Soln:
 The last place of the even number has only two choices (2 or 4) and remaining other 4 places have respectively 4, 3, 2 and 1 choices. 
 Now, by the basic principle of counting, no. of ways forming even numbers = 2 × 4 × 3 × 2 × 1 = 48.

9. How many numbers of different digits between 4000 and 5000 can be formed with the digits 2, 3, 4, 5, 6, 7? 
Soln:
The number between 4000 and 5000 has 4 digits starting with 4 only. 
 So, the thousand place has only 1 choice (i.e. 4 only) and remaining three places has respectively 5, 4 and 3 choices. 
 Now, by the basic principle of counting, required no. of ways = 1 × 5 × 4 × 3 = 60 

10. How many numbers of three different digits can be formed form the integers 2, 3, 4, 5, 6? How many of them will be divisible by 5? 

Soln: 
To form three digit numbers from the given digit. Hundreds place have 5, tens place have 4 and ones place have 3 choices. 
 Now, by the basic principle of counting, total no. of ways to form 3-digit number = 5 × 4 × 3 = 60 Again, for the numbers divisible by 5, the last digit of the numbers should be either 0 or 5. 
 Therefore ones place have 1 choice and remaining two places has 4 and 3 choices. 
 Now, by the basic principle of counting, required no. of ways = 1× 4 × 3 = 12 ways.