Permutations/Arrangements



The number of permutations with "n" different things taking "r" at a time is given by
^{n}P_{r}, where ^{n}P_{r} 
= 

(Or) 
= 
(n)! × (n − 1)! × (n − 2)! × .... "r" times. 

Permutations/Arrangements taking only some of the letters



The number of words that can be formed using all the letters of a "n" letter word taking "r" letters where all the letters are different is given by
"The number of permutations of "n" letters taking "r" at a time"
i.e. ^{n}P_{r} (= n × (n − 1) × (n − 2) × ... "r" times)

Permutations/Arrangements taking all the letters



The number of words that can be formed using all the letters of a "n" letter word taking all the letters ("r" = "n") where all the letters are different is given by
"The number of permutations of "n" letters" i.e. ^{n}P_{n} (= n!)
^{n}P_{n} 
= 

⇒ ^{n}P_{n} 
= 

⇒ ^{n}P_{n} 
= 

⇒ ^{n}P_{n} 
= 
n! 
Explanation
Let us assume that there are "n" places into which "n" letters of a word are arranged.
Since they are to be rearranged, they are removed from their places and are being replaced again.
Now, there are "n" letters and they are to be placed in the "n" places available.





1^{st}

2^{nd}

...

...

n^{th}

The total event of placing the "n" letters in the "n" places an be divided into the subevents of placing each letter starting from the first.
 Total Event (E) = Arranging the "n" letters in "n" spaces
 1^{st} subevent (E1) = Arranging a letter in the First place
 2^{nd} subevent (E2) = Arranging a letter in the Second place
 ...
 ...
 n^{th} subevent (En) = Arranging a letter in the n^{th} place
In arranging the letters,
 The 1^{st} place is filled first with any one of the available "n" letters
which can be done is "n" ways ⇒ n_{E1} = n.
 The 2^{nd} place is filled with any one of the (n − 1) letters remaining after filling the 1^{st} place
which can be done is "n − 1" ways ⇒ n_{E2} = n − 1.
 The 3^{rd} place is filled with any one of the (n − 2) letters remaining after filling the 1^{st} and 2^{nd} places
which can be done is "n − 2" ways ⇒ n_{E3} = n − 2.
 ...
 ...
 The n^{th} place is filled with the 1 letter (n − (n − 1) remaining after filling the first (n − 1) places
which can be done is "1" way ⇒ n_{En} = 1.
Therefore, the no. of ways in which the "n" letters can be filled in the "n" places is given by
n_{E} 
= 
n_{E1} × n_{E2} × n_{E3} × .... × n_{En} 

= 
(n) × (n − 1) × (n − 2) × ... × (1) 

= 
n! 
Illustration
The no. of words that can be formed with the letters of the word "Algebra"
No. of letters in the word "Algebra" = 7 {A, L, G, E, B, R, A}
⇒ No. of words that can be formed using all (n) the letters of the word algebra taking (r) all at a time
= ^{n}P_{r} where n = 7 and r = 7 (Or) n! where n = 7
= ^{7}P_{7} or 7!
= 7!

Fixing letters (each in its own place)



In working out the no. of possible words that can be formed using the letters of an "n" letter word by fixing a letter in a particular place, assume that the total event is divided into two subevents.
 Total Event (E) » Arranging the "n" letters in "n" spaces
 1^{st} subevent (E1) » Arranging the fixed letter in the required place
This can be done in one way
 2^{nd} subevent (E2) » Arranging the remaining "n − 1" letters in the remaining "n − 1" places
This can be done in (n − 1)! ways
Therefore, the no. of ways in which the "n" letters can be filled in the "n" places is given by

n_{E} 
= 
n_{E1} × n_{E2} 


= 
1 × (n − 1)! 


= 
(n − 1)! 
Illustration
The no. of words that can be formed with the letters of the word "Tuesday" such that "T" occupies the middle place
Hide/Show Solution
In the word "Tuesday"
No. of letters = 7 {T, U, E, S, D, A, Y} ⇒ No. of places = 7
 Total Event (E) » Filling the 7 places with the 7 letters
 1^{st} subevent (E1) » Filling the middle place with "T"
No. of ways in which this can be accomplished

⇒ n_{E1} 
= 
^{1}P_{1} (Or) 1! 


= 
1 
 2^{nd} subevent (E2) » Filling the remaining 6 places with the remaining 6 letters
Number of Places/Letters remaining

= 
Total Places/Letters − Specified Letters/Specified Places 

= 
7 − 1 

= 
6 
No. of ways in which this can be accomplished 
⇒ n_{E2} 
= 
^{6}P_{6} (Or) 6! 


= 
6 × 5 × 4 × 3 × 2 × 1 


= 
720 
The number of words that can be formed with the letters of the word tuesday that start with "T"
⇒ n_{E} 
= 
n_{E1} × n_{E2} 

= 
1 × 720 

= 
720 

Two or More Letters
The no. of words that can be formed by
 Fixing two letters in two places (each in its own place) is given by (n − 2)!
 Total Event (E) » Arranging the "n" letters in "n" spaces
 1^{st} subevent (E1) » Arranging the first letter to be fixed in the required place
This can be done in one way
 2^{nd} subevent (E2) » Arranging the second letter to be fixed in the required place
This can be done in one way
 3^{rd} subevent (E3) » Arranging the remaining "n − 2" letters in the remaining "n − 2" places
This can be done in (n − 2)! ways
Therefore, the no. of ways in which the "n" letters can be filled in the "n" places is given by

n_{E} 
= 
n_{E1} × n_{E2} × n_{E3} 


= 
1 × × (n − 2)! 


= 
(n − 2)! 
 Fixing three letters in three places (each in its own place) is given by (n − 3)!
 ...
 ...
 Fixing "n" letters in "n" places (each in its own place) is given by (n − n)! i.e. 0! =1.
[If we fix each letter in a place, then there is only one word we can form with such a restriction.]
Illustration
The no. of words that can be formed with the letters of the word "Thursday" such that "T" occupies the first place and "Y" occupies the last place
Hide/Show Solution
In the word "Thursday"
No. of letters = 8 {T, H, U, R, S, D, A, Y}
No. of places = 8
 Total Event (E) » Filling the 8 places with the 8 letters
 1^{st} subevent (E1) » Filling the first place with "T"
No. of ways in which this can be accomplished 
⇒ n_{E1} 
= 
^{1}P_{1} (Or) 1! 


= 
1 
 2^{nd} subevent (E2) » Filling the last place with "Y"
No. of ways in which this can be accomplished 
⇒ n_{E2} 
= 
^{1}P_{1} (Or) 1! 


= 
1 
 3^{rd} subevent (E3) » Filling the remaining 6 places with the remaining 6 letters
Number of Places/Letters remaining 
= 
Total Places/Letters − Specified Letters/Specified Places 

= 
8 − (1 + 1) 

= 
8 − 2 

= 
6 
No. of ways in which this can be accomplished 
⇒ n_{E3} 
= 
^{6}P_{6} (Or) 6! 


= 
6 × 5 × 4 × 3 × 2 × 1 


= 
720 
The number of words that can be formed with the letters of the word thursday that start with "T" and end with "Y"
⇒ n_{E} 
= 
n_{E1} × n_{E2} × n_{E3} 

= 
1 × 1 × 720 

= 
720 


Fixing a set of (two or more letters) in a set of places



For finding the no. of possible words that can be formed using the letters of an "n" letter word by arranging some of the letters in specific places, assume that the total event is divided into two subevents.
 Total Event (E) » Arranging the "n" letters in "n" spaces
 1^{st} subevent (E1) » Arranging the specified letters in the specified places.
Where "a" represents the number of specified letters and "b" the number of specified places,
The number of ways this can be done (i.e. the number of possibilities)
 2^{nd} subevent (E2) » Arranging the remaining letters in the remaining places.
The remaining letters and the remaining places would be the same in all the above cases
Remaining places/letters 
= 
Total Places/Letters − Lower of Specified Letters/Specified Places 
⇒ n^{c} 
= 
n − a [Where a < b ] 
(Or) 
= 
n − b [Where b < a ] 
(Or) 
= 
(n − a) or (n − b) [Where a = b ] 
Therefore, the number of words that can be formed is given by n_{E} = n_{E1} × n_{E2}
 When a > b, n_{E} = ^{a}P_{b} × (n − b)!
 When b > a, n_{E} = ^{b}P_{a} × (n − a)!
 When a = b, n_{E} = a! (Or) b! × (n − a)! (Or) (n − b)!
Illustrations
1. 
The no. of words that can be formed with the letters of the word "Equation" such that the even spaces are occupied by vowels
Hide/Show Solution
In the word "Equation", the
No. of Letters = 8 {E, Q, U, A, T, I, O, N}
No. of Vowels = 5 {E, U, A, I, O}
No. of Even places = 4 {X, _, X, _, X, _, X, _}
No. of Specified letters (a) = 5 and No. of Specified places (b) = 4
 Total Event (E) » Filling the 8 places with the 8 letters
 1^{st} subevent (E1) » Filling the 4 even places with the 5 letters
No. of ways in which this can be accomplished 
⇒ n_{E1} 
= 
^{5}P_{4} 


= 
5 × 4 × 3 × 2 


= 
120 
 2^{nd} subevent (E2) » Filling the remaining 4 places with the remaining 4 letters
No. of Places/Letters remaining 
= 
Total Places/Letters − Lower of Specified Letters/Specified Places 

= 
8 − 4 

= 
4 
No. of ways in which this can be accomplished 
⇒ n_{E2} 
= 
^{4}P_{4} (Or) 4! 


= 
4 × 3 × 2 × 1 


= 
24 
The no. of words that can be formed with the letters of the word equation such that even places are occupied by vowels.
⇒ n_{E} 
= 
n_{E1} × n_{E2} 

= 
120 × 24 

= 
2,880 


2. 
The letters of the word FAILURE are arranged at random. The number of words that can be formed with the consonants occupying odd positions.
Hide/Show Solution
In the word "Failure", the
No. of letters = 7 {F, A, I, L, U, R, E}
No. of consonants = 3 {F, L, R}
No. of odd positions = 4 {_,X, _, X, _, X, _, X}
No. of specified letters (a) = 3 and No. of specified places (b) = 4
 Total Event (E) » Filling the 7 places with the 7 letters
 1^{st} subevent (E1) » Filling the 3 letters in the 4 odd positions
No. of ways in which this can be accomplished 
⇒ n_{E1} 
= 
^{4}P_{3} 


= 
4 × 3 × 2 


= 
24 
 2^{nd} subevent (E2) » Filling the remaining 4 places with the remaining 4 letters
No. of Places/Letters remaining 
= 
Total Places/Letters − Lower of Specified Letters/Specified Places 

= 
7 − 3 

= 
4 
No. of ways in which this can be accomplished 
⇒ n_{E2} 
= 
^{4}P_{4} (Or) 4! 


= 
4 × 3 × 2 × 1 


= 
24 
The no. of words that can be formed with the letters of the word Failure such that the consonants occupy odd positions
⇒ n_{E} 
= 
n_{E1} × n_{E2} 

= 
24 × 24 

= 
576 


Where there are two or more specification, each specification is to be treated as a sub event.

Two or more letters grouped (stay together)



For finding the no. of possible words that can be formed using the letters of an "n" letter word such that two or more letters stay together, assume that the total event is divided into two subevents.
 Total Event (E) » Arranging the "n" letters in "n" spaces
 1^{st} subevent (E1) » Arranging ((n − a) + 1) letters in as many places consider the letters to be grouped as a unit,
Where "a" represents the number of letters to be grouped together.
The number of ways this can be done ⇒ ^{((n − a) + 1)}P_{((n − a) + 1)} (Or) ((n − a) + 1)!
 2^{nd} subevent (E2) » Arranging the letters which stay together among themselves.
The number of ways this can be done ⇒ ^{a}P_{a} (Or) a!
The no. of words that can be formed 
⇒ n_{E} 
= 
n_{E1} × n_{E2} 


= 
((n − a) + 1)! × a! 
Eg: 
1. 
The no. of words that can be formed with the letters of the word "Victory" such that all the vowels come together
Hide/Show Solution
In the word "Victory" , the
No. of letters = 7 {V, I, C, T, O, R, Y}
No. of vowels = 2 {I, O}
Taking the vowels as a unit
No. of letters = 6 {V, (I,O), C, T, R, Y}
No. of places to be filled = 6
 Total Event (E) » Filling the 7 places with the 7 letters
 1^{st} subevent (E1) » Filling the 6 places with the 6 letters by taking the vowels as group
No. of ways in which this can be accomplished 
⇒ n_{E1} 
= 
^{6}P_{6} (Or) 6! 


= 
6 × 5 × 4 × 3 × 2 × 1 


= 
720 
 2^{nd} subevent (E2) » Inter arranging the two vowels among themselves
No. of ways in which this can be accomplished 
⇒ n_{E2} 
= 
^{2}P_{2} (Or) 2! 


= 
2 times; 1 


= 
2 
The no. of words that can be formed with the letters of the word equation such that all the vowels are together
⇒ n_{E} 
= 
n_{E1} × n_{E2} 

= 
720 × 2 

= 
1,440 



Where there are two or more groups, each group is to be treated as a separate entity
Eg: 
1. 
The no. of words that can be formed with the letters of the word "Daughter" such that all the vowels come together and "DGH" stay together
Hide/Show Solution
In the word "Daughter", the
No. of letters = 8 {D, A, U, G, H, T, E, R}
No. of vowels (first group) = 3 {A, U, E}
No. of vowels (second group) = 3 {D, G, H}
Taking the vowels as a unit and "DGH" as a unit
No. of letters = 4 {(A, U, E), (D, G, H), T, R}
No. of places to be filled = 4
 Total Event (E) » Filling the 4 places with the 4 letters
 1^{st} subevent (E1) » Filling the 4 places with the 4 letters by taking the vowels as group and "DGH" as a group
No. of ways in which this can be accomplished 
⇒ n_{E1} 
= 
^{4}P_{4} (Or) 4! 


= 
4 × 3 × 2 × 1 


= 
24 
 2^{nd} subevent (E2) » Inter arranging the two vowels among themselves
No. of ways in which this can be accomplished 
⇒ n_{E2} 
= 
^{3}P_{3} (Or) 3! 


= 
3 times; 2 times; 1 


= 
6 
 3^{rd} subevent (E3) » Inter arranging the three letters "DGH" among themselves
No. of ways in which this can be accomplished 
⇒ n_{E3} 
= 
^{3}P_{3} (Or) 3! 


= 
3 times; 2 times; 1 


= 
6 
The no. of words that can be formed with the letters of the word daughter such that all the vowels are together and the letters "DGH" are together
⇒ n_{E} 
= 
n_{E1} × n_{E2} × n_{E3} 

= 
24 × 6 × 6 

= 
864 




