Permutation and Combination - Important Points
Permutation and Combination - MCQ
Factorial
Let n be a positive integer. Then n factorial (n!) can be defined as
n! = n (n-1) (n-2)...1
Examples:
6! = 6 x 5 x 4 x 3 x 2 x 1 = 720
3! = 3 x 2 x 1 = 6
Special Cases
0! = 1
1! = 1
Fundamental Principles of Counting - Multiplication Theorem
If an operation can be performed in m different ways and following which a second operation can be performed in n different ways, then the two operations in succession can be performed in m × n different ways.
Example 1: In a class of 5 girls and 4 boys, the teacher has to select 1 girl AND 1 boy. In how many ways can a teacher make the selection?
Solution: Here the teacher has to choose the pair of a girl AND a boy
For 1st boy ------- any one of the 5 girls ----------- 5 ways
For 2nd boy ------- any one of the 5 girls ----------- 5 ways
For 3rd boy ------- any one of the 5 girls ----------- 5 ways
For 4th boy ------- any one of the 5 girls ----------- 5 ways
Total number of ways 5 + 5 + 5 + 5 = 20 ways OR 5 X 4 = 20 ways.
Fundamental Principles of Counting - Addition Theorem
If an operation can be performed in m different ways and a second independent operation can be performed in n different ways, either of the two operations can be performed in (m + n) ways.
Example 2: In a class of 5 girls and 4 boys, the teacher has to select either a girl OR a boy. In how many ways can he make his selection?
Solution: Here the teacher has to choose either a girl OR a boy (Only 1 student)
For selecting a boy he has 4 options/ways OR that for a girl 5 options/ways. The first of these can be performed in 4 ways and the second in 5 ways.
Therefore, by fundamental principle of addition either of the two jobs can be performed in (4 + 5) ways. Hence, the teacher can make the selection of a student in 9 ways.
Example 3: There are 3 candidates for sports, 5 for a mathematical, and 4 for a science scholarship.
I. In how many ways can these scholarships be awarded?
II. In how many ways can one of these scholarships be awarded?
Solution: Clearly, sports scholarship can be awarded to anyone of the 3 candidates. Similarly, mathematical and science scholarship can be awarded in 5 and 4 ways respectively. So,
I. Number of ways of awarding three scholarships = 3 X 5 X 4 = 60 [By Fundamental Principle of Multiplication]
II. Number of ways of awarding one of the three scholarships = 3 + 5 + 4 = 12 [By Fundamental Principle of Addition]
Example 4: A room has 6 doors. In how many ways can a person enter the room through one door and come out through a different door?
Solution: Number of ways coming in the room = 6
Number of ways going out of the room = 5 (He / She cannot go from the same door)
By Fundamental Principle of Multiplication--------> Coming in X Going out = 6 X 5 = 30.
Example 5: Five persons entered the lift cabin on the ground floor of an 8 floor building. Suppose each of them can leave the cabin independently at any floor beginning with the first. Find the total number of ways in which each of the five persons can leave the cabin
(i) At any one of the 7 floors
(ii) At different floors.
Solution: Let the five persons be a, b, c, d, e
(i) a can leave the cabin at any of the seven floors. So he has 7 options. Similarly, each of b, c, d, e also has 7 options. Thus the total number of ways in which each of the five persons can leave the cabin at any of the seven floors = 7 X 7 X 7 X 7 X7 = 7^5
(ii) a can leave the cabin in 7 ways. b can leave the cabin in 6 ways, since he cannot leave at where a left. In the same way c has 5, d has 4, and e has 3 ways.
Hence total number of ways = 7 X 6 X 5 X 4 X 3 = 2520