- Written By P_C
- Last Modified 25-01-2023
Permutations: Suppose you have three letters \(A,B,\) and \(C,\) we can arrange them as \(ABC,ACB,BCA,BAC,CBA,\) and \(CAB\) by taking all the letters simultaneously. We will be able to make more arrangements if using only one or two of the letters if allowed. Here, note that each arrangement is different from the other. So, in other words, the order of writing the letters is critical. So, permutations are generally the arrangement of objects in a linear manner, where the objects may or may not be repeated.
In this article, let us more about permutations of \(n\) distinct objects taken some or all at a time with repetition or without repetition.
What is a Permutation?
The permutation is an arrangement of objects in a definite order.
Example: If there are \(3\) objects let’s say \(a,b,c\) then the permutations of these objects, taking \(2\) at a time, are \(ab,ba,bc,cb,ac,ca\)
So, the number of permutations of \(3\) different objects taken \(2\) at a time is \(6.\)
Note: The order of arrangement is vital in calculating the number of permutations. When the order is changed, a different permutation is obtained.
Factorial Notation
The continuous product of first \(n\) natural numbers is called \(n\) factorial. It is denoted by \(n!\).
\(n! = 1 \times 2 \times 3 \times 4 \times … \times \left( {n – 1} \right) \times n\)
It can also be represented as,
Thus,
\(1! = 1\)
\(2! = 1 \times 2 = 2\)
\(3! = 1 \times 2 \times 3 = 6\)
\(4! = 1 \times 2 \times 3 \times 4 = 24\)
\(5! = 1 \times 2 \times 3 \times 4 \times 5 = 120\)
\(6! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 = 720\)
\(7! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 = 5040\)
\(8! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 = 40320\)
\(9! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 = 362880\)
\(10! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10 = 3628800\)
Clearly, \(n!\) is defined for positive integers only, and it is not defined for negative numbers or fractions.
Representation of a Permutation
If \(n\) and \(r\) are two positive integers such that \(1 \leqslant r \leqslant n,\) then the number of permutations of \(n\) different objects taken \(r\) at a time is
\(P\left( {n,r} \right)\) or \(^n{P_r}\)
Thus, \(P\left( {n,r} \right)\) or \(^n{P_r}\) gives the total number of permutations of \(n\) distinct objects taken \(r\) at a time.
Theorem 1
Statement:
The total number of permutations of \(n\) different or distinct objects taken \(r\) at a time, where \(0 < r \leqslant n\) and the objects do not repeat is \(n\left( {n - 1} \right)\left( {n - 2} \right) \ldots \left( {n - r + 1} \right),\) which is represented by \(^n{P_r}.\)
Proof:
The number of permutations of \(n\) distinct objects taken \(r\) at a time is the same as the number of ways in which we can fill up \(r\)-places when we have \(n\) different objects at our disposal.
1. The first place can be filled in \(n\) ways, for any one of the \(n\) objects can be used.
2. Then, there are \(\left( {n – 1} \right)\) objects left, and any one of these can be used to fill up the second place. Hence, the second place can be filled in \(\left( {n – 1} \right)\) ways.
Hence, by using the fundamental principle of counting, the first two places can be filled in \(n\left( {n – 1} \right)\) ways.
3. Now, there are \(\left( {n – 2} \right)\) places left. The third place can be filled in \(\left( {n – 2} \right)\) ways.
Therefore, the first three places can be filled in \(n\left( {n – 1} \right)\left( {n – 2} \right)\) ways.
4. Similarly, we find that the first \(\left( {r – 1} \right)\) places can be filled in \(n\left( {n – 1} \right)\left( {n – 2} \right)…\left( {n – \left( {r – 2} \right)} \right)\) ways.
5. After filling up first \(\left( {r – 1} \right)\) places, exactly \(n – \left( {r – 1} \right) = n – r + 1\) objects are left. So, the \({r^{{\text{th}}}}\) place can be filled in \(\left( {n – \left( {r – 1} \right)} \right)\) ways. Hence, the \(r\) places can be filled in \(n\left( {n – 1} \right)\left( {n – 2} \right)…\left( {n – \left( {r – 1} \right)} \right)\) ways.
6. Total number of permutations of \(n\) distinct objects, taken \(r\) at a time is \(n\left( {n – 1} \right)\left( {n – 2} \right)\left( {n – 3} \right)…\left( {n – \left( {r – 1} \right)} \right)\)
Thus, \(P\left( {n,r} \right) = n\left( {n – 1} \right)\left( {n – 2} \right)\left( {n – 3} \right) \ldots \left( {n – \left( {r – 1} \right)} \right)\)
Formula for Permutations
The formula that is used to calculate the number of permutations is
\(^n{P_r} = \frac{{n!}}{{\left( {n – r} \right)!}}\)
Here,
\(n\) is the total number of objects, and \(r\) is the chosen objects.
Derivation of Formula
We know that: \(P\left( {n,r} \right) = n\left( {n – 1} \right)\left( {n – 2} \right)\left( {n – 3} \right) \ldots \left( {n – \left( {r – 1} \right)} \right)\)
\( \Rightarrow P\left( {n,r} \right) = \frac{{n\left( {n – 1} \right)\left( {n – 2} \right)\left( {n – 3} \right) \ldots \left( {n – \left( {r – 1} \right)} \right)\left( {n – r} \right)\left( {n – \left( {r + 1} \right)} \right) \ldots 3 \times 2 \times 1}}{{\left( {n – r} \right)\left( {n – \left( {r + 1} \right)} \right) \ldots 3 \times 2 \times 1}}\)
\(\therefore P\left( {n,r} \right) = \frac{{n!}}{{\left( {n – r} \right)!}}\)
Theorem 2
The number of all permutations of \(n\) distinct objects, taken all at a time, is \(n!\)
Proof: The number of all permutations of \(n\) distinct objects, taken all at a time, is the same as the number of ways of filling \(n\) places when we have \(n\) distinct objects at our disposal.
Using Theorem \(1,\) we have
\(P\left( {n,n} \right) = n\left( {n – 1} \right)\left( {n – 2} \right)\left( {n – 3} \right) \ldots \left( {n – \left( {n – 1} \right)} \right)\)
\( = n\left( {n – 1} \right)\left( {n – 2} \right)\left( {n – 3} \right) \ldots 3.2.1\)
\( = n!\)
Theorem 3
The number of permutations of \(n\) different objects taken \(r\) at a time where repetition is allowed is \({n^r}.\)
Proof: When the number of objects is \(n,\) and we have \(r\) to be the selection of an object, then there are \(n\) different ways (each time) to choose an object.
So, the permutation of the objects when the repetition is allowed will be equal to
\(n \times n \times n \times \ldots \left( {r\,{\text{times}}} \right) = {n^r}\)
Types of Permutations
The permutations can be classified into three different categories such as;
1. Permutations of \(n\) distinct objects (when repetition is not allowed)
2. Permutations of \(n\) distinct objects (when repetition is allowed)
3. Permutations when all the objects are not different or distinct
Let us now discuss three categories in detail.
Permutation of Distinct Objects, No Repetition
For suppose \(n\) and \(r\) are two positive integers, such that \(r < n,\) then \(P\left( {n,r} \right)\) represents the number of all possible arrangements or permutations of \(n\) distinct objects taken \(r\) at a time. In the case of permutation without repetition, the number of available choices will be reduced each time. It can also be represented as \({}^n{P_r}\)
Therefore, \(P\left( {n,r} \right) = n\left( {n – 1} \right)\left( {n – 2} \right)\left( {n – 3} \right) \ldots \) up to \(r\) factors
\( \Rightarrow P\left( {n,r} \right) = n\left( {n – 1} \right)\left( {n – 2} \right)\left( {n – 3} \right) \ldots \ldots \left( {n – r + 1} \right)\)
\(\therefore {}^n{P_r} = \frac{{n!}}{{\left( {n – r} \right)!}}\)
Here, \({}^nP\) represents the number of permutations of \(n\) different objects taken \(r\) at a time.
Permutations of Distinct Objects with Repetition
If \(n\) and \(r\) are two positive integers such that \(r < n,\) then \(P\left( {n,r} \right)\) represents the number of possible arrangements or permutations of \(n\) distinct objects taken \(r\) at a time. In the case of permutation with repetition,
The number of permutations of \(n\) different objects taken \(r\) at a time when repetition is allowed is \({n^r}.\)
Permutations When all Objects are not Distinct
Theorem 4: The number of permutations of \(n\) objects, where \(p\) objects are of the same kind and rest all are different \( = \frac{{n!}}{{p!}}\)
Theorem 5
The number of permutations of \(n\) objects, where \({p_1}\) objects are of one kind, \({p_2}\) objects are of the second kind, … \({p_k}\) are of \({k^{th}}\) kind and the remaining all are of a different kind, is given by the formula :
\(\frac{{n!}}{{{p_1}!{p_2}! \ldots {p_k}!}}\)
Solved Examples – Permutations
Q.1. How many ways can two different rings be worn in four fingers with at most one in each finger?
Ans: Required number of ways \( = {}^4{P_2}\)
\( = \frac{{4!}}{{\left( {4 – 2} \right)!}}\)
\( = \frac{{4!}}{{2!}}\)
\( = 12\)
Q.2. Seven athletes are participating in a race. In how many ways can the first two prizes be won?
Ans: Required number of ways \( = {}^7{P_2}\)
\( = \frac{{7!}}{{\left( {7 – 2} \right)!}}\)
\( = \frac{{7!}}{{5!}}\)
\( = \frac{{7 \times 6 \times 5!}}{{5!}}\)
\( = 42\)
Q.3. In how many ways six children can be arranged in a line, such that, two particular children of them are always together.
Ans: Given: Two particular children need to be together, so we consider them as a single object.
Thus, the five objects, including the single object, can be arranged in \(5!\) Ways i.e., \(120.\)
Also, the two children can be arranged in \(2!\) ways.
Hence, the total number of arrangements \(= 5! \times 2! = 120 \times 2 = 240\) ways.
Q.4. Find the number of permutations of the letters in the word ALLAHABAD.
Ans: The word ALLAHABAD has \(9\) letters or objects of which there are \(4\;A’s,2L\;’s\) and rest are all different.
Therefore, the number of arrangements \( = \frac{{9!}}{{4!2!}}\)
\( = \frac{{5 \times 6 \times 7 \times 8 \times 9}}{2}\)
\( = \frac{{15120}}{2}\)
\( = 7560\)
Q.5. How many numbers lying between \(100\) and \(1000\) can be formed with the digits \(1,2,3,4,5\) if the repetition of digits is not allowed?
Ans: Every number lying between \(100\) and \(1000\) is a three-digit number. Therefore, we have to find the number of permutations of five digits \(1,2,3,4,5\) taken three at a time.
Hence, the required number of numbers \( = {}^5{P_3}\)
\( = \frac{5}{{\left( {5 – 3} \right)!}}\)
\( = \frac{{5!}}{{2!}}\)
\( = 5 \times 4 \times 3\)
\( = 60\)
Summary
The permutation is an arrangement of objects in a particular order. While dealing the permutation, one should be concerned about the selection as well as arrangement. In other words, the permutations of \(n\) different objects taken \(r\) at a time is denoted by \( {}^n{P_r}.\) The formula that is used to calculate the number of permutations is \({}^n{P_r} = \frac{{n!}}{{\left( {n – r} \right)!}}.\) There are different types of permutations such as permutation of \(n\) different objects when repetition is not allowed, repetition is allowed, and permutations when all the objects are not distinct. The number of permutations of \(n\) objects, where \(p\) objects are of the same kind and rest all are different \( = \frac{{n!}}{{p!}}.\)
Frequently Asked Questions (FAQs)
Q.1. How do you calculate permutations?
Ans: The number of permutations of \(n\) different objects taken \(r\) at a time is calculated by the formula
\(^n{P_r} = P\left( {n,r} \right) = \frac{{n!}}{{\left( {n – r} \right)!}}\)
Q.2. How many permutations can be made out of the letters of the word TRIANGLE?
Ans: There are eight letters in the given word TRIANGLE
The total number of ways permutations are to put these eight letters in eight places
\(^8{P_8} = \frac{{8!}}{{0!}}\)
\( = 8!\)
\(8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40,320\)
Q.3. How many permutations can be formed by the letters of the word VOWELS?
Ans: There are six letters in the given word VOWELS
The total number of ways permutations are to put these six letters in six places
\(^6{P_6} = \frac{{6!}}{{0!}} = 6! = 720\)
Q.4. What is the permutation with repetition?
Ans: The number of permutations of \(n\) different objects taken \(r\) at a time where repetition is allowed is \({n^r}.\)
Q.5. What are the types of permutations?
Ans: The permutations of an arrangement of objects in order depends on three conditions:
(i) When the repetition of the objects is not allowed.
(ii) When the repetition of the objects is allowed.
(iii) When all the objects are not distinct.
Learn About Permutation Here
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