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Permutation and the combination is the study of arranging elements in a set, and of combining and rearranging elements. Permutations may be represented by a specific sequence of permutation numbers, while combinations may be represented by a specific sequence of combination numbers.
Contents
Concept of Permutation And Combination
Permutation and Combination are the methods of counting which help us to determine the number of different ways of arranging and selecting objects out of a given number of objects, without actually listing them. In this article, you will be able to learn the daily life application of permutation and combination along with their proper meaning and formula.
Before discussing permutation and combination, we will learn the important mathematical term ‘Factorial’.
What is Factorial?
Factorial is defined as the product of all natural numbers less than or equal to a given natural number.
For a natural number ‘n’, its factorial is denoted by n!
And, n! = n (n-1) (n-2) (n-3) (n-4) …… 3 x 2 x 1.
For example: 7! = 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040
10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 3628800
Note: (1) 0! = 1
(2) 1! = 1
What is Permutation?
The arrangement of objects in a definite order is called Permutation.
For example, You have two books, one book on each subject, Mathematics, and Science. You want to keep them properly on a shelf. So, you can keep them either Mathematics book first and then science book (i.e. M than S), or Science book first and then Mathematics book next to it (i.e. S then M). Therefore, there are two ways to arrange the two books on a shelf.
The number of permutations of n different objects taken r objects out of them without replacement, where 0 < r ≤ n, is given by:
\[^{n}P_{r} = \frac {n!}{(n-r)!}\]
From the previous example, the number of permutations of 2 different books, Mathematics and Science, taken both of them is:
\[^{2}P_{2} = \frac {2!}{(2-2)!} = \frac {2!}{0!} = 2! = 2 \times 1 = 2 ways\]
What is Combination?
The selection of some or all objects from a given set of different objects where the order of selection is not considered is called Combination.
For example: Suppose you have three friends, Aman, Mohit, and Raj, you want two of them to go with you for a picnic. You are not able to decide which two of them you should take for a picnic, so you think of an idea that you will write each of their names on a separate paper and pick two of them. Then, the possible ways you could get the names of your two friends will be:
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Aman and Mohit
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Mohit and Raj
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Aman and Raj,
Therefore, there are three ways to select two friends out of three friends.
The number of combinations of n different objects taken r objects out of them without replacement, where 0 < r ≤ n, is given by:
\[^{n}C_{r} = \frac {n!}{r!(n-r)!} = \frac {_{n}Pr}{r!}\]
From the previous example, the number of ways of picking the names of two friends out of three names is:
\[^{3}C_{2} = \frac {3!}{2!(3-2)!} = \frac {3!}{2! \times 1!} =\frac {3 \times 2 \times 1}{2 \times 1} = 3ways\]
Note:
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If in a problem statement, you are asked for selection and their ordering, then you should use Permutation.
In simple words, Permutation = Selection + ordering
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If in a problem statement, you are asked only for selection then you should use Combination.
In simple words, Combination = Selection
Difference Between Permutation and Combination
Solved Examples:
Q.1. How many words can be formed from the letters of the word ‘VEDANTU’ using each letter only once?
Solution: There are 7 letters in the word ‘VEDANTU’.The number of ways of arranging all the 7 letters of a given word is:
\[^{7}P_{7} = \frac {7!}{(7-7)!} = \frac {7!}{0!} = 7! = 5040 words\]
Q.2. How many 5 digit telephone numbers can be formed if each number starts with 21 and no digit appears more than once?
Solution: Since, the first two places have to be filled only by 2 and 1 respectively there is only 1 way for doing this.
Also, it is given that no digit appears more than once. Hence, there will be 8 digits remaining (0, 3, 4, 5, 6, 7, 8, 9) and 3 places have to be filled with these remaining digits.
So, the next 3 places can be filled with the remaining 8 digits in 8P3 ways.
Total number of ways =
\[^{8}P_{3} = \frac {8!}{(8-3)!} = \frac {8!}{5!} = \frac {8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{5 \times 4 \times 3 \times 2 \times 1} = 8 \times 7 \times 6 = 336 \]
Q.3. How many chords can be drawn through 15 points on a circle?
Solution: Since a chord is drawn by joining two points on a circle.
Given that, there are 15 points on a circle. The total number of ways of selecting two points on a circle will give the total number of chords of the circle.
Therefore,the required number of chords
\[^{15}C_{2} = \frac {15!}{2!(15-2)!} = \frac {15!}{2!(15-2)!} = \frac {15!}{2!(13)!} = \frac {15 \times 14}{2 \times 1} = 105\]
Conclusion
This is all about the introduction to permutation and combination. Focus on how this mathematical operation is being conducted and utilized in different subjects. Your conceptual development in this topic will help you get through the advanced subjects of science and mathematics.
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