A permutation is an arrangement of all or part of a number of things in a definite order. For example, the permutations of the three letters a, b, c taken all at a time are abc, acb, bac, bca, cba, cab. The permutations of the three letters a, b, c taken two at a time are ab, ac, ba, bc, ca, cb.
Number of Permutations of n different things taken r at a time:
If n and r are positive integers such that 1≤r≤n, then the number of all permutations of n distinct things, taken r at a time is denoted by the symbol P (n, r) or .
when r = n,
Thus denotes the numbers of permutations of 8 different things taken 3 at a time, and denotes the number of permutations of 5 different things taken 3 at a time.
In permutations, the order of arrangement of elements is taken into account; when the order is changed, a different permutation is obtained.
Example: Find the value of and .
Example: Find the total number of ways in which 4 persons can take their places in a cab having 6 seats.
Solution: The number of ways in which 4 persons can take their places in a cab having 6 seats: = ways.
1. The total number of permutations of n different things taken all at a time is n!.
Example: In how many ways 6 people can stand in a queue?
Solution: The number of ways in which 6 people can stand in a queue:
2. The total number of arrangements of n different things taken r at a time, in which a particular things always occurs =
Example: How many 4 digits number (repetition is not allowed) can be made by using digits 1-7 if 4 will always be there in the number?
Solution: Total digits (n) = 7
Now, as 4 will always occur, so remaining 3 digits could be selected from remaining 6 numbers in .
As in this arrangement, 4 will always occur, so 4 could be placed at any of the 4 places of 4 digit number. So total numbers of ways are 4 x 120 = 480.
This is same as above formula and could also be derived from the above formula as well. Total ways of making the number if 4 is always there =
3. The total number of permutations of n different things taken r at a time in which a particular thing never occurs = .
Example: How many different 3 letters words can be made by 5 vowels, if vowel ‘A’ will never be included?
Solution: Total letters (n) = 5
So total number of ways =
4. The total number of permutations of n dissimilar things taken r at a time with repetitions = .
Example: How many 3 digits number can be made by using digits 1 to 7 if repetition is allowed?
Solution: Total digits (n) = 7
So total ways = .
The number of permutations of n things taken all at a time when p of them are alike and of one kind, q of them are alike and of second kind, all other being different, is:
The above theorem can be extended if in addition to the above r things are alike and of third kind and so on, then
Total permutation =
Example: Find the number of permutations of the letters of the word ASSASSINATION.
Solution: We have 13 letters in all of which 3 are A’s, 4 are S’s, 2 are I’s and 2 are N’s
Therefore the number of arrangements =
Example: Find the number of permutations of the letters of the words ‘DADDY DID A DEADLY DEED’.
Solution: We have 19 letters in all of which 9 are Ds; 3 are As; 2 are Ys; 3 are Es and the rest are all distinct.
Therefore, the number of arrangements =
Example: How many different words can be formed with the letters of word ‘ORDINATE’?
- So that the vowels occupy odd places.
- Beginning with ‘O’.
- Beginning with ‘O’ and ending with ‘E’.
1. ORDINATE contains 8 letters: 4 odd places, 4 vowels.
Number of arrangements of the vowels 4!.
Number of arranging consonants is 4!.
Number of words = 4! x 4! = (4*3*2*1)^2 = 576
2. When O is fixed we have only seven letters at our disposal.
Number of words = 7!=5040
3. When we have only six letters at our disposal, leaving ‘O’ and ‘E’ which are fixed. Number of permutations = 6!=720.
So far we have discussed permutation of objects (or things) in a row. This type of permutations is generally known as linear permutations. If we arrange the objects along a closed curve viz. a circle, the permutations are known as circular permutations. As we have seen in the earlier sections of this chapter that every linear arrangement has a beginning and an end, but there is nothing like a beginning or an end in a circular permutation. Thus, in a circular permutation, we consider one object as fixed and the remaining objects are arranged as in case of linear arrangements.
The number of ways of arranging n distinct objects around a circle is (n -1)! & the total number when taken r at a time will be .
In circular permutation, anti-clockwise and clockwise order or arrangements are considered as distinct permutations.
Example: In how many ways can 7 Australians sit down at a round table?
Solution: 7 Australians can take their seat at the round table in (7-1)! = 6! ways.
Example: In how many ways can 12 persons among whom two are brothers be arranged along a circle so that there is exactly one person between the two brothers?
Solution: One person between the two brothers can be chosen in ways.
The remaining 9 persons can be arranged in .
The two brothers can be arranged in 2! ways.
Therefore, the total number of ways = (9!) (10)( 2!) = (10!)(2!).
If there be no difference between clockwise and anticlockwise arrangements, the total no. of circular permutations of n things taken all at a time is and the total number when taken r at a time will be .
Example: In how many ways can be garland of 10 different flowers be made?
Solution: Since, there is no difference between clockwise and anticlockwise arrangements, the total no. of ways
1. The total number of ways of answering 5 objective type questions, each question having 4 choices is:
2. The expression would be represented by
- P(n, n)
3. Which of the below expression would be equal to P(n,0)
- P(n, 1)
- P(n, n/2)
4. A necklace is a circular string with several beads on it. It is allowed to rotate a necklace but not to turn it over. How many different necklaces can be made using 13 different beads?
5. Which of the below expression would be equal to P (n, r)
- C(n, 0)
- r! C(n, n-r)
- P(n, n-r)
- P(n, n/2)
6. P(n, n) equals
7. The number of ways of distributing 10 different books among 4 students (S1, S2, S3, S4) such that S1 and S2 get 2 books each and S3 and S4 get 3 books each is:
- P(n, r)
9. P(5,r)=5, r equals
10. How many signals can be made with 5 different flags by raising them any number at a time?
11. P(r,2)=56, r equals
12. P(7,3)+ 3P(7,2) equals
13. Find the number of words of 5 letters such that each can be formed with the letters of the word “CHROMATE” if each letter may be repeated in any arrangement.
14. P(11,5)+ 5P(11,4) equals
15. P(13,3) + 3P(13,2) = P(14,r) then r equals
Rubavathi Manoharan says
Answers please. ….
NO. 13 ANSWER?