Permutation
A permutation is an arrangement of objects in a definite order. Since we have already studied combinations, we can also interpret Permutations as ‘ordered combinations’.
Analysis
Let’s say we have a set of ‘n’ distinct objects, out of which we must choose ‘r’ objects. We can proceed in the following two ways of choosing the required objects:
• When the repetition of objects is allowed
• When the repetition of objects is not allowed
Case 1
When the repetition of objects is allowed, at every step of our choosing an object from the set of ‘n’ objects, we have all the ‘n’ choices available to us since we can make a choice multiple times. So, for choosing ‘r’ objects, we have n choices available to us ‘r’ times. Let us call the event of choosing an object as E:
n(E) = n (the number of ways in which E can take place)
Since this event is taking place ‘r’ times and the act of choosing an object from the available set is always independent of our other choices, we may invoke the Product Rule of Counting here. Using the fundamental principle then, we get,
n(E taking place ‘r’ times) = nr
This is the permutation formula for the number of permutations possible for the choice of ‘r’ objects from a set of ‘n’ distinct objects when repetition is allowed.
Case 2
In this case, when the repetition of objects is not allowed, we must be careful, not to choose a specific object more than once. Hence our choices after each event get reduced by one. For example, when we begin choosing our first object, we have all the ‘n’ choices available to us.
In the next event, however, we have ‘(n1)’ objects available for choice, since we must not include the object that we have already chosen in the first step. Similarly, for the third step, we have ‘(n2)’ objects available to us. Thus, from the Product Rule of Counting, we can get,
n(E) = n × (n – 1) × (n – 1) … (n – (r – 1))
(notation for the number of permutations of r objects out of a set of n distinct objects)
A Specific Case
The number of permutations of ‘n’ objects from a set of ‘n’ distinct objects would be given by n!. For example, the number of ways in which you can jumble the alphabets of the word ‘flower’ is given by 6!. What if our reservoir set of ‘n’ objects has some repeated elements?
The formula can be derived in a similar manner to our derivation of the general formula, but with some important restrictions. You could try working it out yourself! But for the time being, let us just state and understand it. The number of permutations of ‘n’ objects where p1 objects are of one kind, p2 objects are of one other kind… till pk , is given by:
where clearly, p1 + p2 +p3 …..+ pk = n.
Example: The number of ways in which you can jumble the alphabets of the word ‘balloon’ is given by
since the number count of different alphabets is given as:
n(occurrence of alphabet ‘b’)= 1
n(occurrence of alphabet ‘a’)= 1
Also, n(occurrence of alphabet ‘l’)= 2
n(occurrence of alphabet ‘o’)= 2
n(occurrence of alphabet ‘n’)= 1
Alternate Analysis of ^{n}P_{r}
If you remember, we had also mentioned ‘ordered combinations’ as another interpretation of ‘permutations’. Let us look at the formula for the number of combinations of ‘r’ objects out of a set of ‘n’ objects:
Let us take a specific combination of these number of different combinations. In permutations, the order of the chosen elements does matter; while in the combinations the order doesn’t matter. Thus, to arrive at the number of permutations, we must multiply the number of combinations by the number of possible permutations of each specific combination!
What is the number of permutations for a set of ‘r’ objects? It is simply r!, as we have discussed some time ago. Multiplying ^{n}C_{r} by r!, we observe:
which is the same as the one we derived from our Product Rule of Counting. Therefore, the interpretation of ‘Permutations’ as ‘Ordered Combinations’ is correct!
Solved Example for You on Permutation Formula
Question: In a sports broadcasting company, the manager must pick the top three goals of the month, from a list of ten goals. In how many ways can the top three goals be decided?
Solution: Since the manager must decide the top three goals of the month; the order of the goals is very important! It decides the firstplace winner, the runnerup, and the second runnerup. Thus, we can see that the problem is of permutation formula.
Picking up three goals from a list of ten:
Possible Permutations = ^{10}P_{3} = = 10 × 9 × 8 = 720
Therefore, there are 720 ways of picking the top three goals!
Fundamental Principles of Counting
1. Multiplication Principle
If first operation can be performed in m ways and then a second operation can be performed in n ways. Then, the two operations taken together can be performed in mn ways. This can be extended to any finite number of operations.
2. Addition Principle
If first operation can be performed in m ways and another operation, which is independent of the first, can be performed in n ways. Then, either of the two operations can be performed in m + n ways. This can be extended to any finite number of exclusive events.
Factorial
For any natural number n, we define factorial as n ! or n = n(n – 1)(n – 2) … 3 x 2 x 1 and 0!= 1!= 1
Permutation
Each of the different arrangement which can be made by taking some or all of a number of things is called a permutation.
Mathematically The number of ways of arranging n distinct objects in a row taking r (0 ≤ r ≤ n) at a time is denoted by P(n ,r) or ^{n}p_{r}
Properties of Permutation
Important Results on’Permutation
Division into Groups
(i) The number of ways in which (m + n) different things can be divided into two groups which contain m and n things respectively [(m + n)!/m ! n !].
This can be extended to (m + n + p) different things divided into three groups of m, n, p things respectively [(m + n + p)!/m!n! p!].
(ii) The number of ways of dividing 2n different elements into two groups of n objects each is [(2n)!/(n!)^{2}] , when the distinction can be made between the groups, i.e., if the order of group is important. This can be extended to 3n different elements into 3 groups is [(3n)!/((n!)^{3}].
(iii) The number of ways of dividing 2n different elements into two groups of n object when no distinction can be made between the groups i.e., order of the group is not important is [(2n)!/2!(n!)^{2}].
This can be extended to 3n different elements into 3 groups is [(3n)!/3!(n!)^{3}].
The number of ways in which mn different things can be divided equally it into m groups, if order of the group is not important is [(mn)!/(n!)^{m }m!].
(v) If the order of the group is important, then number of ways of dividing mn different things equally into m distinct groups is mn [(mn)!/(n!)^{m}]
(vi) The number of ways of dividing n different things into r groups is [r^{n} — ^{r}C_{1}(r — 1)^{n} + ^{r}C_{2}(r — 2)^{n} — ^{r}C_{3}(r – 3)^{n} + …].
(vii) The number of ways of dividing n different things into r groups taking into account the order of the groups and also the order of things in each group is ^{n+r1}P_{n} = r(r + l)(r + 2) … (r + n – 1).
(viii) The number of ways of dividing n identical things among r persons such that each gets 1, 2, 3, … or k things is the coefficient of x^{n – r} in the expansion of (1 + x + x^{2} + … + X^{k1})^{r}.
Circular Permutation
In a circular permutation, firstly we fix the position of one of the objects and then arrange the other objects in all possible ways.
(i) Number of circular permutations at a time is (n 1)!. If clockwise taken as different. of n and different things taken anticlockwise orders all are
(ii) Number of circular permutations of n different things taken all at a time, when clockwise or anticlockwise order is not different 1/2(n – 1)!.
(iii) Number of circular permutations of n different things taken r at a time, when clockwise or anticlockwise orders are take as different is ^{n}P_{r}/r
(iv) Number of circular permutations of n different things taken r at a time, when clockwise or anticlockwise orders are not different is ^{n}P_{r}/2r.
(v) If we mark numbers 1 to n on chairs in a round table, then n persons sitting around table is n!.
1. What is a permutation? 
2. How do you calculate the number of permutations? 
3. What is the difference between a permutation and a combination? 
4. How can permutations be used in reallife situations? 
5. Can permutations be used to solve probability problems? 
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157 videos210 docs132 tests
