Chapter : 1
Basic Principles Of Counting | Class 12 Math | Algebra | Notes and Solutions |Important Questions
Basic principle of counting is the part of permutation and combination. In this article you will learn about how many methods of counting and its important questions with solutions . For more study materials follow Edubook Ram Kumar Sah.
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Fundamental Principles Of Counting
a. Addition Principle
(a) Addition principle: If an event occur in m different ways and a second event occur in n different ways. then either of the two events can occur in (m+n) ways provided only one event can occur at a time.
Suppose A and B be two cities. There are 3 bus routes and 2 train routes between A and B. A person can go from city 4 to city B by any one of the 3 bus routes or any one of the 2 train routes. Thus, there are 3+2=5 ways by which the person can travel from city A to city B. In this case, the simultaneous use of the bus route and train route is not possible. This principle implies addition
(b) Multiplication principle
(b) Multiplication principle: If an even can occur in m different ways and if corresponding to each way of occurring this event there are n different ways of occurrence of the second event, then both the events can occur simultaneously in m x n different ways."
Let A, B and C are three cities. Between A and B, there are 3 roads and between B and C. there are 2 roads. A person Rajesh staying at A want to go to city C. For this, Rajesh has to pass through city B. In going from A to B, Rajesh has 3 choices to choose the road. In the same way, he has 2 choices to choose the road in going from B to C. Each way of choosing the road between A and B is associated with 2 ways of choosing the road between B and C. If the roads between A and B are denoted by 1, 2, 3 and the roads between B and C by a, b, then the different ways of choosing the roads from A to C are (1. a), (1, b), (2, a), (2, b), (3, a), (3, b) i.e. 6 ways.
Thus, the total number of ways by which Rajesh can go from A to C = 3x2=6.
Hence, if there are n ways to reach from A to B and ways to reach from B to C, then their simultaneous occurrence will be mn ways. This rule is known as the basic principle of counting. This rule can be extended to three or more events. This rule implies multiplication.
-Worked Out Examples--
1. In a cinema-hall, there are three entrance doors and two exit doors. In how many ways can a person enter the hall and then come out.
Solution:
Clearly, a person can enter the hall through any of the three entrance door. So, there are 3 ways of entering the hall. After entering the hall, the person can come out through any of the two exit doors. So, there are 2 ways of coming out. Hence, the number of ways in which a person can enter the hall and then come out is 3 x 2 = 6.
2. How many numbers of three different digits can be formed from the integers 2, 3, 4, 5, 6?
Solution:
To form a number of three different digits, we are to fill up three places: the hundred's, the ten's and the unit's. HTU
The hundred's place can be filled in 5 different ways, because any one of 2, 3, 4, 5, 6 can fill up this place. The ten's place can be filled up by remaining digits in 4 ways. The unit place can be filled up by 3 ways.
∴ The required numbers of three different digits = 5 x 4 x 3 = 60.
3. There are 3 candidates for a social science, 5 for a mathematical science, and 4 for a natural science scholarship.
i. In how many ways can these scholarships be awarded?
ii. In how many ways one of these scholarships is awarded?
Solution:
i. Clearly social science scholarship can be awarded to any one of the three candidates. So, there are 3 ways of awarding the social science scholarship. Mathematical science scholarship can be awarded in 5 ways and natural science scholarship can be awarded in 4 ways. So, the number of ways of awarding three scholarships =3x5x4=60.
ii. Number of ways in which one of these scholarships is awarded =3+5+4=12.
4. There are 6 doors in a hostel. In how many ways can a boy enter the hostel and leave by the different door?
Solution:
As there are 6 doors in a hostel, the number of ways the boy can enter the hostel = 6.
As, he is not allowed to leave the hostel by the same door,
so the number of ways the boy can leave the hostel = 5.
By the Basic Principle of Counting, total number of ways = 6 x 5 = 30.
5. The flag of newly formed Club is of the form of three blocks, each to be colured differently If there are six different colours on the whole to choose from, how many such designs are possible?
Solution:
Since there are six colours to choose from, therefore, the first block can be coloured in 6 ways. Also, the second block can be coloured by any one of the remaining five coloures in 5 ways.
After colouring first two blocks only four colours are left. So, there are 4 ways to colour third block.
Hence, by the fundamental principle of multiplication, the number of flag designs =6x5x4=120 ways.
6. How many words (with and without meaning) of three distinct letters of the English alphabet are there?
Solution:
We have to fill up three places by using the letters of English alphabet. Since there are 26 letters in English alphabet: the first place can be filled by any of these letters. So, there are 26 way of filling up the first place. Now, the second place can be filled up by any one of remaining 25 letters. So, there are 25 ways of filling up the second place. After filling up the first two places, only 24 letters are left to fill up the third place. So, the third place can be filled in 24 ways.
Hence, the total number of ways of forming words = 26 x 25 x 24 = 15600.
7. There are 6 multiple choice questions in an examination. How many sequence of answer are possible, if the first three questions have 4 choice each and the next three have 5 each?
Solution:
Here, we have to perform 6 jobs of answering 6 multiple choice questions. Each one of the first three questions can be answered in 4 different ways and each one of the next three can be answered in 5 different ways. So, the total number of ways of answering the questions =4×4×4×5× 5 × 5 = 8000.
8. Given the digits 4, 5, 6, 7, 8, 9, how many numbers of
a. Three different digits can be formed?
b. Three digits can be formed (repetition of digits is allowed.
c. At least three different digit can be formed?
As there are 6 numbers given for the digits, the number of choices for hundred's place = 6.
As the number has to be different digits, the number of choice for ten's place = 5 and the number of choices for unit's place = 4.
Hence, the total number of ways = 6 x 5 x 4 = 120
c. The required number is of three or more digits. Hence, we have to count the numbers of three, four, five and six different digits. Following the above procedures.
Number of three different digits = 6 * 5 * 4 = 120
Number of four different digits=6x5x4x3=360
Number of five different digits = 6 * 5 * 4 * 3 * 2 = 720
Number of six different digits = 6 * 5 * 4 * 3 * 2 * 1 = 720
So, the total number of at least three different digits 120+360+ 720 + 720 = 1920.
9. Find how many odd numbers of four different digits can be formed from the digits 1, 2, 3, 4, 5, 6.
Whether the number is odd or even it is determined by the digit at the unit's place. To form the four digits odd numbers, the choices for the units place are 1 or 3 or 5. We can place any number in other places. Hence,
the number of choices for units place = 3,
the number of choices for ten's place = 5,
the number of choices for hundred's place = 4 and
the number of choices for thousand's place = 3.
Total numbers = 3 * 5 * 4 * 3 = 180 .
10. How many numbers of three different digits less than 600 can be formed by using the digits 2, 3, 4, 6, 72
Solution:
Three digits number less than 600 will have 2, 3 or 4 at hundred's place. So, Number of ways in which hundred's place can be filled up = 3
Number of ways in which ten's place can be filled up = (5 - 1) = 4
Number of ways in which units place can be filled up = (4 - 1) = 3
Hence, the total number of ways forming required numbers = 3 * 4 * 3 = 36 .
11. How many four different digit numbers can be formed using the digits 0, 1, 2, 3, 4, 5?
EXERCISE 1.1 (A) Class 12 Math
1. Define the fundamental principle of addition and multiplication with examples.
2. There are 20 books of Mathematics and 18 books of Physics in a bookshop. In how many ways can a student buy?
(a) A Mathematics book and a Physics book?
(b) Either a Mathematics book or a Physics book?
3. Three persons enter a room where there are seven vacant chairs in a line, find in how many ways they can take their seats?
4. Suppose there are five main roads between cities A and B and 4 between cities B and C. In how many ways can a person drive from A to C and return without driving on the same road twice?
5. How many three digit numbers can be formed without repetation using digits 0, 2, 3, 4, 5 and 6?
6. How many numbers are there between 100 and 1000 in which all the digits are distinct?
7. How many numbers are there between 100 and 1000 such that every digit is either 2 or 9?
8. How many numbers of three different digits less than 500 can be formed from the integers 1,2,3,4, 5, 6?
9. How many numbers between 4000 and 5000 can be formed without repetition with the digit 2, 3, 4, 5, 6, 7?
10. How many numbers can be formed from the integers 1, 2, 3, 4, 5, 6? If
(a) At least three different digits can be formed.
(b) At most three different digits can be formed.
11. By using the figures 1, 2, 3, 4, 5 only once, how many are even numbers.
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