Permutation Computation
Permutation means arrangement of things. The word arrangement indicates that the order of things is considered.
In mathematics, Permutation refers to the act of arranging all the members of a set into some sequence or order, or if the set is already ordered, rearranging (reordering) its elements
For example, the anagram of a word having all different letters, is permutation of letters. In the original word, the letters are already ordered.
For the set {1,2,3}, there are six possible permutation, namely: (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), and (3,2,1)
Permutation is the number of ways of arranging or selecting smaller or equal number of persons or objects from a group of persons or collection of objects considering the order of arrangement or selection.
Factorial
The factorial function n (written as |n or n! ), represents the product of the series of integers from 1 to n both inclusive.
Thus, n! = n(n − 1) (n − 2) ………..3.2.1
Characteristics of Factorial
i) 0!=1, ii) n! is defined only for positive integers, iii) n!=n ´ (n-1)!
Ex. 1: 3! = 3 X 2 X 1 = 6 ; Ex. 2: 4! = 4 X 3 X 2 X 1 = 24; Ex 3: 6! = 6 X 5 X 4 X 3 X 2 X 1= 720.
Ex 4.Compute (32!) / (28!) = {(32x31x30x29)} x28!] / 28! = (32x31x30x29) = 863040.
Permutation Formula
The number of possible arrangements are is denoted by npr, where P stands for permutation, n is the given number of things and ‘r’ is the number of things taken at a time.
For example 5p2 denotes number of permutation or arrangement of 5 objects taken 2 at a time.
![\displaystyle ^{n}{{P}_{r}}=\left[ {\frac{{\left( {n!} \right)}}{{(n-r)!}}} \right]](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5E%7Bn%7D%7B%7BP%7D_%7Br%7D%7D%3D%5Cleft%5B+%7B%5Cfrac%7B%7B%5Cleft%28+%7Bn%21%7D+%5Cright%29%7D%7D%7B%7B%28n-r%29%21%7D%7D%7D+%5Cright%5D&bg=ffffff&fg=6a6c6e&s=0 "\displaystyle ^{n}{{P}_{r}}=\left[ {\frac{{\left( {n!} \right)}}{{(n-r)!}}} \right]")
Circular Permutation
When the objects are arranged in a closed curve (no starting or no end), such arrangement or permutation is known as circular permutation.
For example, for 4 items a,b,c,d, the arrangement abcd, dabc, cdab, bcda are same in circular combination, where there is no start or end positions. So, these 4 linear permutations are equivalent to 1 in circular permutation. Hence, there are nPn /n ways or (n-1)! ways in circular permutation. This is like fix up one position and then arrange the rest (n-1) in (n-1)! ways.
Ex: A Board meeting is attended by 6 directors sitting in chairs in a circle. In how many ways can they arrange themselves ?
Fix up one position and arrange the remaining five about him. Possible number of arrangements 5! = 5 x 4 x 3 x 2. = 120 ways
Permutation with Restrictions
Rule 1 : Number of Permutation of n distinct objects taken r at a time, when a particular object is not taken is n-1Pr.
Since the particular object is always excluded, we have n-1 distinct tings to choose from. So, the permutations are n-1Pr.
Rule 2: Number of Permutation of n distinct objects taken r at a time, when a particular object is always taken is r(n-1Pr-1).
If the particular object is first placed, from remaining n-1 objects, r-1 objects can be selected in n-1Pr-1 ways. Similarly, placing the particular object in 2nd, 3rd … upto rth place, we have n-1Pr-1 Permutations. So, the total permutations are r( n-1Pr-1)
Permutation – Problems
Permutation – Problems and Solution
Ex 1: Two executives enter in a meeting room where there are seven chairs in a line, find in how many ways can they take their seats.
Number of ways = nPr . Here n=7, r=2, So, the number of permutation
Ex.2 : In how many ways can the letters of the word DAUGHTER be arranged so that the vowels may never be separated ?
As vowels must be together (never separated), consider the 3 vowels (AUE) as one letter. Considering the 3 vowels as one letter, we have only 6 letters D,G,H,T,R (AUE) to arrange. Here the number of arrangements keeping the vowels together = 6P6 = 6! = (6x5x4x3x2) = 720
Again the vowels (while keeping them together), ay be arranged in 3P3 ways, i.e in 3! ways, i.e in (3×2) ways, i.e in 6 ways
So, combining both the permutations, the word can be arranged in 720 x 6 = 4320 ways
Ex. 3: How many different words can be made out of the letters of the word MATHEMATICS? In how many of these will the vowels always come together?
Step 1 : Number of different words: There are 11 letters in the word MATHEMATICS. In non-vowels (consonants), M, A, T are repeating twice. So, the possible permutations are
![\displaystyle \left[ {\frac{{\left( {11!} \right)}}{{\left( {2!\times 2!\times 2!} \right)}}} \right]=\frac{{\left( {11!} \right)}}{8}](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cleft%5B+%7B%5Cfrac%7B%7B%5Cleft%28+%7B11%21%7D+%5Cright%29%7D%7D%7B%7B%5Cleft%28+%7B2%21%5Ctimes+2%21%5Ctimes+2%21%7D+%5Cright%29%7D%7D%7D+%5Cright%5D%3D%5Cfrac%7B%7B%5Cleft%28+%7B11%21%7D+%5Cright%29%7D%7D%7B8%7D&bg=ffffff&fg=6a6c6e&s=0 "\displaystyle \left[ {\frac{{\left( {11!} \right)}}{{\left( {2!\times 2!\times 2!} \right)}}} \right]=\frac{{\left( {11!} \right)}}{8}")
= 39916800/8= 49,89,600
Step 2 : Number of different words with vowels together : The number of permutations in which vowels would come together
![\displaystyle =\text{ }\left[ {8!\div \left( {2!\text{ }2!} \right)} \right]\times \left( {\frac{{4!}}{2}} \right)](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%3D%5Ctext%7B+%7D%5Cleft%5B+%7B8%21%5Cdiv+%5Cleft%28+%7B2%21%5Ctext%7B+%7D2%21%7D+%5Cright%29%7D+%5Cright%5D%5Ctimes+%5Cleft%28+%7B%5Cfrac%7B%7B4%21%7D%7D%7B2%7D%7D+%5Cright%29&bg=ffffff&fg=6a6c6e&s=0 "\displaystyle =\text{ }\left[ {8!\div \left( {2!\text{ }2!} \right)} \right]\times \left( {\frac{{4!}}{2}} \right)")
![\displaystyle =\left[ {\frac{{\left( {8\times 7!} \right)}}{{\left( {2\times 2\times 2} \right)}}} \right]\times \left[ {\frac{{\left( {4!} \right)}}{{2x}}} \right]](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%3D%5Cleft%5B+%7B%5Cfrac%7B%7B%5Cleft%28+%7B8%5Ctimes+7%21%7D+%5Cright%29%7D%7D%7B%7B%5Cleft%28+%7B2%5Ctimes+2%5Ctimes+2%7D+%5Cright%29%7D%7D%7D+%5Cright%5D%5Ctimes+%5Cleft%5B+%7B%5Cfrac%7B%7B%5Cleft%28+%7B4%21%7D+%5Cright%29%7D%7D%7B%7B2x%7D%7D%7D+%5Cright%5D&bg=ffffff&fg=6a6c6e&s=0 "\displaystyle =\left[ {\frac{{\left( {8\times 7!} \right)}}{{\left( {2\times 2\times 2} \right)}}} \right]\times \left[ {\frac{{\left( {4!} \right)}}{{2x}}} \right]")
= (7!)x (4!) = 5040 x 24=1,20,960 (There are 11 letters, 8 distinct and 3 repeating twice. There are 4 vowels, with repeating 2 A’s)
Permutation – Problems
Permutation – Problems and Solution
Ex 1: Six boys and five girls are to be seated for a photograph in a row such that no two girls sit together and no two boys sit together. Find the number of ways in which this can be done.
Step 1 : We have 11 persons in a row and we want the 6 boys and 5 girls to be seated such that no two girls and no two boys are together. If we number the positions from left to right, the arrangement will be possible if and only if boys occupy the odd places and girls occupy the even places in the row.
Step 2 : The six odd places from 1 to 11 may filled in by 6 boys in 6P6 ways.
Five even places from 2 to 10 may be filled in by 5 girls in 5P5 ways.
Step 3 : Hence, the total number of required arrangements = 6P6 X 5P5 =
6! X 5! = 720 X 120 = 86400.
Permutation – Problems
Permutation – Problems and Solution
Ex 1:In how many ways can the letter of the word ‘balloon’ be arranged, so that two l’s do not come together ?
There are 7 letters in the word balloon, of which the letter l and o are repeated 2 times each
So the no. of arrangements without restriction 
Considering 2 l’s as one letter, we have only 6 letters, arrangement may be done in (6!) / (2!) = 360 ways
So, the reqd. arrangement, in which 2 l’s do not come together = 1260 – 360 = 900.
Combination
Combination is the number of ways in which smaller or equal number of things are arranged or selected from a collection of things, where the order of selection or arrangement is not Important
The expression of Combination is denoted by nCr where C is the number of combination, n is number of objects, r is the number of things taken at a time nCr = [(n!) / {r! x (n-r)!}]
Ex 12C4 = [12! / {(12-4) ! x 4!}] = [(12!) / {(8!) X (4!)}] = [{(12x11x10x9x8!)} / {(8!) x (4!)}]
= (12x11x10x9) / (4x3x2) = 495
Characteristics of Combination
(i) nCn =1, (ii) nC0 = 1, (iii) nCr =nCn–r , (iv) nC1 = n (v) n+1Cr=ncr+nCr-1
Combination – Problems
Combination – Problems and Solution
Ex 1: A committee of 7 members is to be chosen from 7 company secretaries (CS), 4 lawyers (L) and 6 Engineers (E). In how many ways can this be done if in the committee to have at least one member from each group and at least 3 company secretaries?
Possible Methods of Combination
Method 1 : 3 CS, 2L, 2 E= 7C3 x4C2 x5C2 ![\displaystyle =\left[ {\frac{{(7\times 6\times 5)}}{{(3\times 2)}}} \right]\times \left( {\frac{{\left( {4\times 3} \right)}}{2}} \right)\text{ }\times ~\left[ {\frac{{\left( {5\times 4} \right)}}{2}} \right]=2100](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%3D%5Cleft%5B+%7B%5Cfrac%7B%7B%287%5Ctimes+6%5Ctimes+5%29%7D%7D%7B%7B%283%5Ctimes+2%29%7D%7D%7D+%5Cright%5D%5Ctimes+%5Cleft%28+%7B%5Cfrac%7B%7B%5Cleft%28+%7B4%5Ctimes+3%7D+%5Cright%29%7D%7D%7B2%7D%7D+%5Cright%29%5Ctext%7B+%7D%5Ctimes+%7E%5Cleft%5B+%7B%5Cfrac%7B%7B%5Cleft%28+%7B5%5Ctimes+4%7D+%5Cright%29%7D%7D%7B2%7D%7D+%5Cright%5D%3D2100&bg=ffffff&fg=6a6c6e&s=0 "\displaystyle =\left[ {\frac{{(7\times 6\times 5)}}{{(3\times 2)}}} \right]\times \left( {\frac{{\left( {4\times 3} \right)}}{2}} \right)\text{ }\times ~\left[ {\frac{{\left( {5\times 4} \right)}}{2}} \right]=2100")
Method 2 : 4 CS, 2L, 1 E= 7C4 x4C2 x5C1 ![\displaystyle =\left[ {\frac{{(7\times 6\times 5)}}{{(3\times 2)}}} \right]\times \left( {\frac{{\left( {4\times 3} \right)}}{2}} \right)\text{ }\times ~\left( 5 \right)=1050](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%3D%5Cleft%5B+%7B%5Cfrac%7B%7B%287%5Ctimes+6%5Ctimes+5%29%7D%7D%7B%7B%283%5Ctimes+2%29%7D%7D%7D+%5Cright%5D%5Ctimes+%5Cleft%28+%7B%5Cfrac%7B%7B%5Cleft%28+%7B4%5Ctimes+3%7D+%5Cright%29%7D%7D%7B2%7D%7D+%5Cright%29%5Ctext%7B+%7D%5Ctimes+%7E%5Cleft%28+5+%5Cright%29%3D1050&bg=ffffff&fg=6a6c6e&s=0 "\displaystyle =\left[ {\frac{{(7\times 6\times 5)}}{{(3\times 2)}}} \right]\times \left( {\frac{{\left( {4\times 3} \right)}}{2}} \right)\text{ }\times ~\left( 5 \right)=1050")
Method 3 : 4 CS, 1L, 2 E= 7C4 x4C1 x5C2 ![\displaystyle =\left[ {\frac{{(7\times 6\times 5)}}{{(3\times 2)}}} \right]\times \left( 4 \right)\times ~\left[ {\frac{{\left( {5\times 4} \right)}}{2}} \right]=1400](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%3D%5Cleft%5B+%7B%5Cfrac%7B%7B%287%5Ctimes+6%5Ctimes+5%29%7D%7D%7B%7B%283%5Ctimes+2%29%7D%7D%7D+%5Cright%5D%5Ctimes+%5Cleft%28+4+%5Cright%29%5Ctimes+%7E%5Cleft%5B+%7B%5Cfrac%7B%7B%5Cleft%28+%7B5%5Ctimes+4%7D+%5Cright%29%7D%7D%7B2%7D%7D+%5Cright%5D%3D1400&bg=ffffff&fg=6a6c6e&s=0 "\displaystyle =\left[ {\frac{{(7\times 6\times 5)}}{{(3\times 2)}}} \right]\times \left( 4 \right)\times ~\left[ {\frac{{\left( {5\times 4} \right)}}{2}} \right]=1400")
Method 4 : 5 CS, 1L, 1 E= 7C5 x4C1 x5C1 
Method 5 : 3 CS, 3L, 1 E= 7C3 x4C3 x5C1 ![\displaystyle =\left[ {\frac{{(7\times 6\times 5)}}{{(3\times 2)}}} \right]\times \left( {\frac{{\left( {4\times 3\times 2} \right)}}{{3\times 2}}} \right)\text{ }\times ~5=700](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%3D%5Cleft%5B+%7B%5Cfrac%7B%7B%287%5Ctimes+6%5Ctimes+5%29%7D%7D%7B%7B%283%5Ctimes+2%29%7D%7D%7D+%5Cright%5D%5Ctimes+%5Cleft%28+%7B%5Cfrac%7B%7B%5Cleft%28+%7B4%5Ctimes+3%5Ctimes+2%7D+%5Cright%29%7D%7D%7B%7B3%5Ctimes+2%7D%7D%7D+%5Cright%29%5Ctext%7B+%7D%5Ctimes+%7E5%3D700&bg=ffffff&fg=6a6c6e&s=0 "\displaystyle =\left[ {\frac{{(7\times 6\times 5)}}{{(3\times 2)}}} \right]\times \left( {\frac{{\left( {4\times 3\times 2} \right)}}{{3\times 2}}} \right)\text{ }\times ~5=700")
Method 6 : 3 CS, 1L, 3 E= 7C3 x4C1 x5C3
Therefore, total number of ways = 2,100 + 1,050 + 1,400 + 420 + 700 + 1,400 = 7070 ways.
Combination – Problems
Combination – Problems and Solution
Ex 1: How many different cricket teams of 7 players may be formed from 8 batsmen and 5 bowlers members ?
Combination of 7 players from 8 Batsmen and 5 Bowlers
Team 1 : Batsmen (7) Bowlers (0) = 8C7 X 5C0 = 8×1 = 8
Team 2 : Batsmen (6) Bowlers (1) = 8C6 X 5C1 = 28×5= 140
Team 3 : Batsmen (5) Bowlers (2) = 8C5 X 5C2 = 560
Team 4 : Batsmen (4) Bowlers (3) = 8C4 X 5C3 = 700
Team 5 : Batsmen (3) Bowlers (4) = 8C3 X 5C4 = 280
Team 6 : Batsmen (2) Bowlers (5) = 8C2 X 5C5 = 28
Different cricket teams can be formed in
( 8 + 140 + 560 + 700 + 280 + 28 ) = 1716 ways
Combination – Problems
Combination – Problems and Solution
Ex 1: From 7 teachers (T) and 4 office staff (S), a group of 5 is to be formed. In how many ways can this be done to include at least one office staff ?
1 office staff can be selected out of 4 office staff in 4C1 ways and 4 teachers can be selected from 7 teachers in 7C4 ways. Now each way of selecting office staff can be associated with each way of selecting teachers. So 1 office staff and 4 teachers can be selected in 4C1 X 7C4 ways.
Combination of a group of 5 of 5 Teachers / Staff
Group 1 : S1 + T4= 4C1 X 7C4 ways = 4×35 = 140
Group 2 : S2 + T3= 4C2 X 7C3 ways = 6x 35 = 210
Group 3 : S3 + T2= 4C3 X 7C2 ways = 4×21 = 84
Group 4 : S4 + T1= 4C4 X 7C1 ways = 1×7 = 7
So, the groups can be formed in 140 + 210 + 84 + 7 = 441 ways
Combination – Problems
Combination – Problems and Solution
Ex.1 If nP4 : nP5=1:4 Compute the value of n.
nP4 : nP5 =1 : 4.
Step 1 : So [{n(n – 1 )(n -2) (n-3 )} / {n(n – 1 )(n – 2) (n – 3 )(n – 4)}] = 1/4
Or , n-4 =4, or n= 4+4 = 8
Ex 2: If nC 10 = nC15 Compute the value of 28Cn.
Step 1: nC 10 = nC14, or nC 10 = nCn-14
So, 10=n-14, n=24
Step 2 : 28Cn =28C24 = 28C28 – 24 =28C4
Ex 3: A firm wishes to simultaneously promote two of its 6 department heads to assistant managers. In how many ways these promotions can take place?
The promotion of candidates do not matter in which order they are done.
So, it can be done in 
Ex. 4 : A Company needs 2 engineers from 9 men applying for the post. In how many ways can these selections take place?
The selection of candidates do not matter in which order they are taken. Hence, the company can select in any of 9C2 ways 
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