Introduction:
Permutation & combination deal with the techniques of counting without direct listing of the number of elements in a particular set or the number of outcomes of a particular experiment. The content of this article may be too rudimentary for most readers, but for beginners, it will be helpful.one has to work hard on basics. Once the basics are very clear, permutation is a very systematic subject.Case 1: “MY FRUIT SALAD IS A COMBINATION OF APPLES, GRAPES & BANANAS”. We don’t care what order the fruits are in, they could be “grapes, apples & bananas” or “bananas, grapes & apples” it’s the same fruit salad.
Case 2: “THE COMBINATION TO THE SAFE LOCK IS 289”.Now we do care about the order. “982” would not work nor would “892”. It has to be exactly 289.
In mathematics we use more accurate language:
 If the order doesn’t matter as in case 1, then it is a COMBINATION
 If the order does matter as in case 2, then it is a PERMUTATION
PERMUTATIONS:
The number of permutations on ‘n’ different things taken ‘r’ at a time is the same as different ways in which ‘r’ places can be filled up with ‘n’ given things. It is denoted by npr or p (n, r) & is given bynPr = n! / (nr)! ………………. (1)
Before going to types of permutations, let us know some basics about solving problems.
1) Find n if np3 = 210
From eqn 1, we can write as,
n! / (n3)! = 210
n (n1)(n2)(n3)!/(n3)! =210
On cancelling (n3) on both numerator & denominator, we get,
n(n1)(n2) = 210
Assume n=7
7x6x5=210
Therefore, n=7 is the ans.
Different types of Permutations:
Type 1
When a certain number of things always occur together
To know better, consider the following given example.
To know better, consider the following given example.
1) In how many ways can 7 people be seated in a row for a photograph, if two particular people always want to be together.
Solution:Consider the two people who want to be together as 1 unit. The number of permutations of remaining 5 people + 1 unit (2 people) = six units is 6!
For each of these two people considered as one unit can be arranged in 2! Ways.
Therefore, Total permutations = 6! X 2!
= 720 X 2 = 1440 ways
For each of these two people considered as one unit can be arranged in 2! Ways.
Therefore, Total permutations = 6! X 2!
= 720 X 2 = 1440 ways
Type 2
When certain number of things occurring together is more than one type
Example:In how many ways 5 quant books , 4 reasoning books & 6 English language books can be arranged, so that if books of the same subject are always together.
Solution:Consider all 5 quant books as 1 unit, 4 reasoning books as 1 unit & 6 English books as 1 unit. So, total 3 units can be arranged in 3!
For each of these quant, reasoning & English books can be arranged in 5! , 4! & 6! Respectively
Total permutations = 3! 4! 5! 6! Or 12,441,600 ways.
Example:In how many ways 5 quant books , 4 reasoning books & 6 English language books can be arranged, so that if books of the same subject are always together.
Solution:Consider all 5 quant books as 1 unit, 4 reasoning books as 1 unit & 6 English books as 1 unit. So, total 3 units can be arranged in 3!
For each of these quant, reasoning & English books can be arranged in 5! , 4! & 6! Respectively
Total permutations = 3! 4! 5! 6! Or 12,441,600 ways.
Type 3
When certain things of one type do not occur together
Example In how many ways can 10 examination papers be arranged so that the best & worst papers never come together.
Solution:Total ways in which best & worst are never together = Total arrangements – Two papers are always together
Total ways = 10!(9! X 2!)
For (9! X 2!) Please refer Type 1 example.
Example In how many ways can 10 examination papers be arranged so that the best & worst papers never come together.
Solution:Total ways in which best & worst are never together = Total arrangements – Two papers are always together
Total ways = 10!(9! X 2!)
For (9! X 2!) Please refer Type 1 example.
Type 4
Formation of numbers with digits (digits can be repeated)
Example How many 3 digit numbers can be formed using digits 1, 2,5,6,8 so that the digits can be repeated.
Example How many 3 digit numbers can be formed using digits 1, 2,5,6,8 so that the digits can be repeated.
Solution:Three digit number consists of 1unit place, 1 tens place &1 hundred place. In the given problem there are 5 numabers, as it can be repeated any no.of times.
Each place can be filled by any one of 5 digits
Total numbers=5x5x5 =125
125 such 3 digits number can be formed.
Each place can be filled by any one of 5 digits
Total numbers=5x5x5 =125
125 such 3 digits number can be formed.
COMBINATION
Type 5
When no two of a certain type occur together
Example: In how many ways can 9 boys & 6 girls be arranged in a row if no two girls are together
Example: In how many ways can 9 boys & 6 girls be arranged in a row if no two girls are together
Solution: If no two girls can be together, we first arrange the boys. This can be done in 9! Ways. In the gaps we have to arrange the girls. Considering the left most & right most positions & the gaps in between the boys. Let ‘B’ be the position of boys & ‘G’ be the position of girls.
G1 B1 G2 B2 G3 B3 G4 B4 G5 B5 G6 B6 G7 B7 G8 B8 G9 B9 G10
Therefore, in 10 places i.e. (from G1 to G10) 6 girls can be arranged in 10P6 ways.
So, total ways = 9! X 10P6
Note:  No two girls are together is different from all the girls are not together. All the girls are not together means few girls can be together.
G1 B1 G2 B2 G3 B3 G4 B4 G5 B5 G6 B6 G7 B7 G8 B8 G9 B9 G10
Therefore, in 10 places i.e. (from G1 to G10) 6 girls can be arranged in 10P6 ways.
So, total ways = 9! X 10P6
Note:  No two girls are together is different from all the girls are not together. All the girls are not together means few girls can be together.
Type 6
Formation of numbers with digits (digits are not repeated) Case1: When there is no restriction
Example:  How many different 4 digit numbers can be formed using the digits 1,2,4,5,7,8,9 no digit being repeated in any number.
Example:  How many different 4 digit numbers can be formed using the digits 1,2,4,5,7,8,9 no digit being repeated in any number.
Solution:
7

6

5

4

The thousand place (Th) can be filled in 7 ways. For each of these the hundred, tenth, units place can be filled in 6, 5, 4 ways respectively
Total ways = 7x6x5x4=840 nos.
Case 2:
Numbers divisible by a particular number or even number or odd numbers
Example:  How many 4 digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 no digit being repeated.
Example:  How many 4 digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 no digit being repeated.
Solution: If the number is even it has to end with even number (2, 4 or 6 in this given question).Hence units place has 3 choices & thousandth , hundredth & tenth place can be filled in 5 ,4,3 ways respectively.
Therefore, total 4 digit even numbers = 5x4x3x3 =180
Therefore, total 4 digit even numbers = 5x4x3x3 =180
Type 7
Word building (Alphabets not repeated)
Case 1: When there is no restriction
Example:  Find the number of words that can be formed by using all the letters of the word “MONKEY”.
Solution: The word MONKEY has 6 letters. The six different letters can be arranged in 6P6 = 6! = 720 ways.
Case 1: When there is no restriction
Example:  Find the number of words that can be formed by using all the letters of the word “MONKEY”.
Solution: The word MONKEY has 6 letters. The six different letters can be arranged in 6P6 = 6! = 720 ways.
Case 2:
When few letters/vowels/consonants occur together
Example: In how many ways can the letters of word “LAUGHTER” be arranged so that the vowels are always together.
Example: In how many ways can the letters of word “LAUGHTER” be arranged so that the vowels are always together.
Solution: The word LAUGHTER has 3 vowels & 5 consonants. Consider the 3 vowels as 1unit.Hence 1unit+ 5 consonants.
Total= 6 can be arranged in 6! Ways. For each of these, the 3 vowels can be arranged in 3! Ways.
Therefore, total words= 6! x3! =720x6=4320
Total= 6 can be arranged in 6! Ways. For each of these, the 3 vowels can be arranged in 3! Ways.
Therefore, total words= 6! x3! =720x6=4320
Case 3:
When vowels/consonants occupy odd/even places
Example: In how many ways can the letters of the word HEXAGON be arranged so that the vowels are always in even places.
Solution: The word HEXAGON has 7 letters of which 3 are vowels. Since total letters is 7, there are 4 odd places & 3 even places.
The 3 vowels can be arranged in the 3 odd places in 3! Ways. For each of these the 4 consonants can be arranged in the 4 odd
Places in 4! Ways. Hence total ways = 3!x4!=6x24=144
Example: In how many ways can the letters of the word HEXAGON be arranged so that the vowels are always in even places.
Solution: The word HEXAGON has 7 letters of which 3 are vowels. Since total letters is 7, there are 4 odd places & 3 even places.
The 3 vowels can be arranged in the 3 odd places in 3! Ways. For each of these the 4 consonants can be arranged in the 4 odd
Places in 4! Ways. Hence total ways = 3!x4!=6x24=144
Type 8
Permutation of like things The number of permutations of ‘n’ things taken all at a time, where ‘x’ of the things are alike & of one kind, ‘y’ others are alike of another kind, ‘z’ others are like & of another kind & so on is
Total No.of ways = n! / x!y!z!
Example:  Find the number of permutations of the letters of the word ENGINEERING. How many of these
a) Begin with E & end with E
b) Have all the 3 E’s together
c) No two vowels are together
Solution: The word ENGINEERING has 11 letters out of which E is repeated thrice, N is repeated thrice, G is repeated twice, I is repeated twice.
Total permutations = 11! /3! 3! 2! 2!
a) Beginning with E & ending with E:
Out of 3 E’s, the position of 2 E’s are fixed i.e, first & last.
Total 9,N3,G2,I2
Hence permutations of remaining letters is 9!/3!2!2!
b) All 3 E’s are together
Consider the 3 E’s as one unit. Hence we have to arrange 1 unit (3 E’s) & remaining letters.
Total No.of ways = n! / x!y!z!
Example:  Find the number of permutations of the letters of the word ENGINEERING. How many of these
a) Begin with E & end with E
b) Have all the 3 E’s together
c) No two vowels are together
Solution: The word ENGINEERING has 11 letters out of which E is repeated thrice, N is repeated thrice, G is repeated twice, I is repeated twice.
Total permutations = 11! /3! 3! 2! 2!
a) Beginning with E & ending with E:
Out of 3 E’s, the position of 2 E’s are fixed i.e, first & last.
Total 9,N3,G2,I2
Hence permutations of remaining letters is 9!/3!2!2!
b) All 3 E’s are together
Consider the 3 E’s as one unit. Hence we have to arrange 1 unit (3 E’s) & remaining letters.
EEE & NGINRING
1 + 8 =9
Total=9, N3, G2, I2
Total permutations = 9! / (3! 2! 2!)
The 3 E’s considered as one can be arranged in only 1 way as all the 3 letters are the same.
Total permutation = 9! / (3! 2! 2!)
c) No two vowels are together
First arrange 6 consonants. N G N R N G
N3, G2, total6
This can be done in 6! / (3! 2!)
If there are 6 letters then there are 7 gaps. In the 7 gaps we have to arrange the 5 vowels. But in these 5 vowels E is repeated thrice & I is repeated twice.
So total ways = 7P3 / (3! 2!)
Total words with no two vowels are together is
{[6! / (3! 2!)] X [7P3 / (3! 2!)]}
1 + 8 =9
Total=9, N3, G2, I2
Total permutations = 9! / (3! 2! 2!)
The 3 E’s considered as one can be arranged in only 1 way as all the 3 letters are the same.
Total permutation = 9! / (3! 2! 2!)
c) No two vowels are together
First arrange 6 consonants. N G N R N G
N3, G2, total6
This can be done in 6! / (3! 2!)
If there are 6 letters then there are 7 gaps. In the 7 gaps we have to arrange the 5 vowels. But in these 5 vowels E is repeated thrice & I is repeated twice.
So total ways = 7P3 / (3! 2!)
Total words with no two vowels are together is
{[6! / (3! 2!)] X [7P3 / (3! 2!)]}
Type 9
Circular Permutations When objects are arranged along a closed curve like a circle or ring, the permutations are known as circular permutations.
It is important to note that in a circular arrangement there is neither a beginning (first term) nor an end (last term)
1) The number of circular permutations of ‘n’ objects is (n1)!
2) If the clockwise & anticlockwise orders are not distinguishable, then the number of circular permutations is (n1)! / 2
Example:
i) In how many ways can 8 persons be seated at a round table?
Solution: This is circular permutation. Hence 8 persons can be seated at a round table is (81)! =7!
ii) In how many ways can 10 different precious stones be set to form necklace
Solution: In case of necklace there is no distinction b/w the clockwise & anticlockwise arrangement. Hence no. of ways = (101)! /2 = 9! /2
It is important to note that in a circular arrangement there is neither a beginning (first term) nor an end (last term)
1) The number of circular permutations of ‘n’ objects is (n1)!
2) If the clockwise & anticlockwise orders are not distinguishable, then the number of circular permutations is (n1)! / 2
Example:
i) In how many ways can 8 persons be seated at a round table?
Solution: This is circular permutation. Hence 8 persons can be seated at a round table is (81)! =7!
ii) In how many ways can 10 different precious stones be set to form necklace
Solution: In case of necklace there is no distinction b/w the clockwise & anticlockwise arrangement. Hence no. of ways = (101)! /2 = 9! /2
Exercise problems for practice
1) Find the value ofa) 12P4
b) 8P6
2) In how many ways can 9 soldiers stand in a queue?
3) How many different signals can be made by taking 3 different coloured flags at a time from 7 different coloured flags?
4) In how many ways can 5 letters be posted in 5 letter boxes if each box has one letter
5) How many six digit numbers can be formed with the digits 2,7,6,1,9,8
6) In how many ways can 4 people occupy 6 vacant chairs
7) In how many ways can 7 persons be seated in a row if two persons always occupy the end seats
8) In how many ways the word “CARROM” be arranged such that the 2 R’s are always together.
9) In how many ways can 7 students & 4 teachers be seated in a row such that no two teachers are together
10) Find the number of permutations of the letters of the word “MATHEMATICS”
11) In how many ways can 10 people be seated around a table
12) Find the number of ways in which 8 men be arranged round a table so that 2 particular men may not be next to each other
2) 9!
3) 210
4) 120
5) 720
6) 360
7) 240
8) 120
9) 7! 8P4
10) 11! / (2! 2! 2!)
11) 9!
12) 7! – (6! 2!)
2) In how many ways can 9 soldiers stand in a queue?
3) How many different signals can be made by taking 3 different coloured flags at a time from 7 different coloured flags?
4) In how many ways can 5 letters be posted in 5 letter boxes if each box has one letter
5) How many six digit numbers can be formed with the digits 2,7,6,1,9,8
6) In how many ways can 4 people occupy 6 vacant chairs
7) In how many ways can 7 persons be seated in a row if two persons always occupy the end seats
8) In how many ways the word “CARROM” be arranged such that the 2 R’s are always together.
9) In how many ways can 7 students & 4 teachers be seated in a row such that no two teachers are together
10) Find the number of permutations of the letters of the word “MATHEMATICS”
11) In how many ways can 10 people be seated around a table
12) Find the number of ways in which 8 men be arranged round a table so that 2 particular men may not be next to each other
Solutions:
1) a) 11880 b) 201602) 9!
3) 210
4) 120
5) 720
6) 360
7) 240
8) 120
9) 7! 8P4
10) 11! / (2! 2! 2!)
11) 9!
12) 7! – (6! 2!)
Smart Prep Kit for Banking Exams by Ramandeep Singh  Download here
0 comments:
Post a Comment
Thanks for commenting. It's very difficult to answer every query here, it's better to post your query on IBPSToday.com