Table of contents |
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Introduction |
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Fundamental Principle of Counting |
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What is Permutation? |
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What is Combination? |
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Circular Permutations |
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MNP Rule |
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Important Results |
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Solved Examples |
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Factorial Notation:
There are some theorems involved in finding the permutations when all the objects are distinct. They are:
Relationship Between Permutation and combination:
In permutation and combination for class 11, the relationship between the two concepts is given by two theorems. They are:
Permutation and combination have various applications in real-life scenarios and different fields of study. Some common applications include:
Example: The numbers 1, 2, 3, 4 and 5 are to be used for making three-digit numbers without repetition In how many ways, can this be done?
Sol: Lets visualize it like we have these three places to fill _____ _____ _____
and we have five numbers i.e 1 , 2 , 3 , 4 , 5 to choose from and we have to fill these places taking into consideration that repetition of the numbers is not allowed.
So, there are 5 different ways to fill the first blank (As we have five different numbers to choose from)
Now, As we moved to the second blank, first blank would already have been filled,
So, No. of ways in which 2nd blank can be filled = 4 ways
similarly, No. of ways in which 3rd blank can be filled = 3 ways
now, the total number of ways in which this three digit number can be made out of these 5 numbers is = 5 x 4 x 3 = 60 ways.
The following results are very important as this will help you in problem solving
Example 1: From a pack of cards in how many ways can you select-
a) A king and a queen
b) A king or a queen
Sol: There are 52 cards in a pack, with 4 kings and 4 queens.A king can be selected in 4 ways and a queen can be selected in 4 ways. Number of ways of selection of a king AND a queen is 4×4=16 (Product rule)
Number of ways of selection of a king OR a queen 4+4=8 (Addition rule)
The fundamental principle of counting (FPC) is the basic concept of permutation and combination.
Example 2: In how many ways can 3 people A, B, C be arranged in 2 places?
Sol: Manually solving we will get 6 cases as given below.AB BA
AC CA
BC CB
This can be explained using FPC like this
There are two seats __ __.
In the first seat A,B or C can sit. i.e., 3 possibilities.Let A sit in the first place. Thus, in second place either B or C can sit.
The second seat can hence be taken up in 2 ways Together A,B ‘AND’ C can be seated in 3 x 2 = 6 ways.
Thus, the number of arrangement of 3 things at 2 places can be done in 3×2 ways.
Example 3: In how many ways can 4 people be seated in 3 chairs?
Sol:
- First seat can be filled in 4 ways
- Second seat can be filled in 3 ways.
- Third seat can be filled in 2 ways.
- So all together 4 people can be arranged in 3 seats in 4x3x2 = 24 ways.
[Question: 475654
Sol:
- Choose the President: There are 3 choices.
- Choose the Vice-President: After choosing the President, there are 2 remaining choices= 3 × 2 = 6
There are 6 ways to select a President and a Vice-President from 3 people.
Example 5: a, b, c are three distinct integers from 2 to 10 (both inclusive). Exactly one of ab, bc and ca is odd. abc is a multiple of 4. The arithmetic mean of a and b is an integer and so is the arithmetic mean of a, b and c. How many such triplets are possible (unordered triplets)?
Sol:
- Exactly one of ab, bc and ca is odd
- Two are odd and one is even.
- abc is a multiple of 4
- Which means the even number is a multiple of 4.
- The arithmetic mean of a and b is an integer implies a and b are odd.
and so is the arithmetic mean of a, b and c.- hence, a + b + c is a multiple of 3.
c can be 4 or 8.- c = 4; a, b can be 3, 5 or 5, 9
- c = 8; a, b can be 3, 7 or 7, 9
- Four triplets are possible.
- Hence the correct answer is 4.
Example 6: If we listed all numbers from 100 to 10,000, how many times would the digit 3 be printed?
Sol:
- 3-Digit Numbers (100 to 999):
- Hundreds place (300-399): 100 times
- Tens place (130-139, 230-239, ..., 930-939): 90 times
- Units place (103, 113, ..., 993): 90 times
- Total for 3-digit numbers:
- 4-Digit Numbers (1,000 to 9,999):
- Thousands place (3000-3999): 1000 times
- Hundreds place (1300-1399, ..., 9300-9399): 900 times
- Tens place (x30, x31, ..., x39 for each thousand): 900 times
- Units place (x03, x13, ..., x93 for each thousand): 900 times
- Total for 4-digit numbers:
1000+ 900 + 900 + 900 = 3700
Total occurrences of '3' from 100 to 9,999:
Example 7: All numbers from 1 to 150 (in decimal system) are written in base 6 notation. How many of these will contain zero's?
Sol:
- Any multiple of 6 will end in a zero. There are 25 such numbers.
- Beyond this, we can have zero as the middle digit of a 3-digit number.
- This will be the case for numbers from 37 - 41, 73 - 77, 109 - 113, and 145 - 149. There are 20 such numbers.
- Overall, there are 45 numbers that have a zero in them.
- Hence the correct answer is 45.
198 videos|165 docs|152 tests
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1. What is the difference between permutation and combination? | ![]() |
2. How do you calculate the number of permutations of n items taken r at a time? | ![]() |
3. What is meant by circular permutations? | ![]() |
4. Can you explain the MNP rule in permutations and combinations? | ![]() |
5. What are some important results related to permutations and combinations? | ![]() |
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