# Arranging Letters of a Word (all letters different)

# Permutations/Arrangements

^{n}P

_{r}, where

^{n}P_{r} | = |
| ||

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

# Permutations/Arrangements in making words from letters

n_{L} : Number of letters with which the words are to be formed

n_{P} : Number of places to fill the letters i.e. the number of letters in the word to be formed

n_{P} ≤ n_{L}

Since the maximum length of the word that can be formed is limited to the number of letters available

Number of n_{P} lettered words that can be formed using n_{L} letters where all the letters are different

⇒ Number of words that can be formed using n_{L} letters taking n_{P} letters at a time

⇒ Number of permutations of n_{L} items taking n_{P} at a time

⇒ ^{nL}P_{nP} | = |
| ||

= | (n_{L}) × (n_{L} − 1) × (n_{L} − 2) × .... n_{P} times |

## Explanation

Forming a n_{P}letter word with n

_{L}letters can be assumed as the act of arranging n

_{L}letters into n

_{P}places.

1^{st} | 2^{nd} | ... | ... | n_{Pth} |

**Total Event (E)**Arranging the n

_{L}letters in n_{P}places

The total event can be divided into the sub-events of placing each letter in a place starting from the first.

**1**^{st}sub-event (SE_{1})Placing a letter in the First place

The 1^{st}place can be filled with any one of the available n_{L}letters⇒ SE

_{1}can be accomplished in n_{L}ways⇒ n

_{SE1}= n_{L}**2**^{nd}sub-event (SE_{2})Placing a letter in the Second place

The 2^{nd}place can be filled with any one of the remaining n_{L}− 1 letters⇒ SE

_{2}can be accomplished in n_{L}− 1 ways⇒ n

_{SE2}= n_{L}− 1**3**^{rd}sub-event (SE_{3})Placing a letter in the Third place

The 3^{rd}place can be filled with any one of the remaining n_{L}− 2 letters⇒ SE

_{3}can be accomplished in n_{L}− 2 ways⇒ n

_{SE3}= n_{L}− 2- ...
- ...
**n**_{Pth}sub-event (SE_{nP})Placing a letter in the n

_{Pth}placeThe n

_{Pth}place can be filled with any one of the n_{L}− (n_{P}− 1) letters remaining after filling the first n_{P}− 1 places⇒ SE

_{3}can be accomplished in n_{L}− (n_{P}− 1) ways⇒ n

_{SEnp}= n_{L}− n_{P}+ 1

By the fundamental counting principle of multiplication,

Number of ways in which the n_{L} letters can be filled in the n_{P} places

⇒ n_{E} | = | n_{SE1} × n_{SE2} × n_{SE3} × .... × n_{SEp} |

= | (n_{L}) × (n_{L} − 1) × (n_{L} − 2) × ... × (n_{L} − n_{P} + 1) = (n ≡ n × (n − 1) × (n − 2) × ... × (n − r + 1) | |

= | ^{nL}P_{nP} ≡ |

## Example

**The number of 5 letter words that can be formed with the letters of the word subdermatoglyphic**

Number of letters

= 17

{S, U, B, D, E, R, M, A, T, O, G, L, Y, P, H, I, C}

⇒ n_{L} = 17

Number of letters in the word to be formed

⇒ Number of places to be filled in forming the word

= 5

⇒ n_{P} = 5

Number of words that can be formed with the letters of the word subdermatoglyphic

= Number of words that can be formed using n_{L} letters taking n_{P} letters at a time

= Number of permutations of n_{L} items taking n_{P} at a time

= ^{nL}P_{nP}

= ^{17}P_{5}

= 17 × 16 × 15 × 14 × 13

## Where all letters are used

Where all letters are used in forming the words, n_{L} = n_{P}

Number of ways in which the n_{L} letters can be filled in the n_{P} places

⇒ n_{E} | = | ^{nL}P_{nP} |

= | ^{nL}P_{nL} | |

= | n_{L}! | |

Or | = | ^{nP}P_{nP} |

= | n_{P}! |

## Example

**The number of words that can be formed with the letters of the word Algebra**

Number of letters

= 7

{A, L, G, E, B, R, A}

⇒ n_{L} = 7

Number of letters in the word to be formed

⇒ Number of places to be filled in forming the word

= 7

⇒ n_{P} = 7

Number of words that can be formed with the letters of the word Algebra

= Number of words that can be formed using n_{L} letters taking n_{P} letters at a time

= Number of permutations of n_{L} items taking n_{P} at a time

= ^{nL}P_{nP}

= n_{L}!

= 7!

= 4,090

# Fixing letters (each in its own place)

n_{L} : Number of letters with which the words are to be formed

n_{P} : Number of places to fill the letters i.e. the number of letters in the word to be formed

n_{P} ≤ n_{L}

Since the maximum length of the word that can be formed is limited to the number of letters available

n_{FL} : Number of letters to be fixed each in its own place

After fixing n_{FL} letters each in its own place

Number of Letters remaining

= Total number of letters − Number of letters to be fixed

⇒ n_{RL} = n_{L} − n_{FL}

Number of Places remaining to be filled

= Total number of places − Number of letters to be fixed

⇒ n_{RP} = n_{P} − n_{FL}

Number of n_{P} letter words that can be formed using n_{L} letters by fixing n_{FL} letters each in its own place is given by

^{nRL}P_{nRP}

## Explanation

**Total Event (E)**Arranging the n

_{L}letters in n_{P}spaces

Assume that the total event is divided into two sub-events.

**1**^{st}sub-event (SE_{1})Arranging the n

_{FL}letters each in its own placeNumber of ways in which this event can be accomplished

= 1

Since each letter can be placed in a specific place only there is only one way this can be done whatever may be the number of letters being fixed

⇒ n

_{SE1}= 1**2**^{nd}sub-event (SE_{2})Filling the remaining n

_{RP}places with the remaining n_{RL}lettersNumber of ways in which this event can be accomplished

= Number of ways in which n _{RP}places can be filled with the n_{RL}lettersOr = Number of permutations or arrangements of n _{RL}items taking n_{RP}items at a time⇒ n _{SE2}= ^{nRL}P_{nRP}

The number of ways in which the n_{L} letters can be filled in the n_{P} places with n_{FL} letters fixed each in its own place

⇒ n_{E} | = | n_{SE1} × n_{SE2} |

= | 1 × ^{nRL}P_{nRP} | |

= | ^{nRL}P_{nRP} |

### Fundamental Counting principle of Multiplication

If a total event can be sub divided into two or more sub events all of which are independent, then the total number of ways in which the total event can be accomplished is given by the product of the number of ways in which each sub event can be accomplished.## Example

Number of 7 letter words that can be formed with the letters of the word incomputably such that T occupies the middle place

Number of letters in the word

= 12

{I, N, C, O, M, P, U, T, A, B, L, Y}

n_{L} = 12

Number of places to be filled in forming the words

= 7

n_{P} = 7

Number of letters fixed each in its own place

n_{FL} = 1

After fixing the specified letters in their respective places

Number of letters remaining

= Total Number of letters − Number of letters fixed in specific places

⇒ n_{RL} | = | n_{L} − n_{FL} |

= | 12 − 1 | |

= | 11 |

Number of places remaining to be filled

= Total Number of Places − Number of letters fixed in specific places

⇒ n_{RP} | = | n_{P} − n_{FL} |

= | 7 − 1 | |

= | 6 |

The number of 7 letter words that can be formed with the letters of the word incomputably such that T occupies the middle place

= Number of ways in which 12 letters can be arranged in 7 places such that 1 letter is fixed in a specified place

= ^{nRL}P_{nRP}

= ^{11}P_{6}

= 11 × 10 × 9 × 8 × 7 × 6

### using fundamental principle

**Total Event (E)**Filling the 7 places with the 12 letters

**1**^{st}sub-event (SE_{1})Filling the middle place with T

Number of ways in which this event can be accomplished

= Number of ways in which the 1 place can be filled with the 1 letter Or = Number of permutations or arrangements with 1 item taking 1 at a time ⇒ n _{SE1}= ^{1}P_{1}= 1! = 1 **2**^{nd}sub-event (SE_{2})Filling the remaining places with the remaining letters

Number of ways in which this event can be accomplished

= Number of ways in which the remaining 6 places can be filled with the remaining 11 letters Or = Number of permutations or arrangements with 11 items taking 6 at a time ⇒ n _{SE2}= ^{11}P_{6}

The number of 7 letter words that can be formed with the letters of the word incomputably such that T occupies the middle place

⇒ n_{E} | = | n_{SE1} × n_{SE2} |

= | 1 × ^{11}P_{6} | |

= | ^{11}P_{6} | |

= | 11 × 10 × 9 × 8 × 7 × 6 |

## Where all letters are used

Where all letters are used in forming the words, n_{L} = n_{P}

⇒ n_{L} − n_{FL} = n_{P} − n_{FL}

⇒ n_{RL} = n_{RP}

Number of n_{P} lettered words that can be formed with the n_{L} letters, fixing n_{FP} letters each in its own place

= | ^{nRL}P_{nRP} | |

= | ^{nRL}P_{nRL} | |

= | n_{RL}! | |

Or | = | ^{nRP}P_{nRP} |

= | n_{RP}! |

## Example

The number 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

Number of letters

= 8

{T, H, U, R, S, D, A, Y}

n_{L} = 8

Number of places to be filled in forming the words

= 8

n_{P} = 8

Number of letters fixed each in its own place

n_{FL} = 2

After fixing the specified letters in their respective places

Number of letters remaining

= Total Number of letters − Number of letters fixed in specific places

⇒ n_{RL} | = | n_{L} − n_{FL} |

= | 8 − 2 | |

= | 6 |

Number of places remaining to be filled

= Total Number of Places − Number of letters fixed in specific places

⇒ n_{RP} | = | n_{P} − n_{FL} |

= | 8 − 2 | |

= | 6 |

Number 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

= Number of ways in which 8 letters can be arranged in 8 places such that 2 letters are fixed each in its specified place

= ^{nRL}P_{nRP}

= ^{nRL}P_{nRL} Or ^{nRP}P_{nRP}

= n_{RL}! or n_{RP}!

= 6!

= 720

### using fundamental principle

**Total Event (E)**Filling the 8 places with the 8 letters

**1**^{st}sub-event (SE_{1})Filling the first place with T and last place with Y

Number of ways in which this event can be accomplished

= Number of ways in which the 2 places can be filled with the 2 specified letters each in its own place

= 1

⇒ n

_{SE1}= 1**2**^{nd}sub-event (SE_{2})Filling the remaining places with the remaining letters

Number of ways in which this event can be accomplished

= Number of ways in which n _{RP}remaining places can be filled with the n_{RL}remaining lettersOr = Number of permutations or arrangements with n _{RP}items taking all at a time⇒ n _{SE2}= ^{nRL}P_{nRP}= ^{6}P_{6}= 6! = 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_{SE1} × n_{SE2} |

= | 1 × 720 | |

= | 720 |

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

n_{L} : Number of letters with which the words are to be formed

n_{P} : Number of places to fill the letters i.e. the number of letters in the word to be formed

n_{P} ≤ n_{L}

The maximum length of the word that can be formed is limited to the number of letters available.

n_{SL} : Number of letters specified to be filled in the specified places

n_{SP} : Number of places specified to be filled with the specified letters

n_{FP} : Number of places filled

If n_{SL} ≠ n_{SP},

Number of places filled would be the lesser of n_{SL} and n_{SP}.

If the specified letters are lesser only that many letters can be used up for filling and if the specified places are lesser only that many places can be filled.

⇒ n_{FP} = smaller of n_{SL} and n_{SP}

Where n_{SL} = n_{SP},

n_{FP} = n_{SP} = n_{SL}

After fixing n_{FL} letters each in its own place

Number of Letters remaining

= Total number of letters − Number of letters to be fixed

⇒ n_{RL} = n_{L} − n_{FL}

Number of Places remaining to be filled

= Total number of places − Number of letters to be fixed

⇒ n_{RP} = n_{P} − n_{FL}

n_{RP} ≤ n_{RL}

The places remaining to be filled cannot exceed the letters remaining to be used

Number of n_{P} letter words that can be formed using n_{L} letters by fixing n_{FL} letters each in its own place is given by

n_{RL}Pn_{RP}

## Explanation

**Total Event (E)**Arranging the n

_{L}letters in n_{L}spaces

Assume that the total event is divided into two sub-events.

**1**^{st}sub-event (SE_{1})Arranging the n

_{SL}specified letters in the n_{SP}specified placesNumber of ways in which this event can be accomplished

= Number of ways in which the n _{SP}specified places can be filled with the n_{SL}specified lettersOr = Number of permutations or arrangements with greater of n _{SP}and n_{SL}items taking the lesser of them at a time⇒ n _{SE1}= ^{a}P_{b}Number of permutations of a items taking b at a time

Where

a = larger of n

_{SL}and n_{SP}and b the otherIf n

_{SL}= n_{SP},a = b = n

_{SL}= n_{SP}**2**^{nd}sub-event (SE_{2})Arranging the remaining letters in the remaining places.

Number of ways in which this event can be accomplished

= Number of ways in which n _{RP}places can be filled with the n_{RL}lettersOr = Number of permutations or arrangements of n _{RL}items taking n_{RP}at a time⇒ n _{SE2}= ^{nRL}P_{nRP}

The number of words that can be formed

⇒ n_{E} | = | n_{SE1} × n_{SE2} |

= | ^{a}P_{b} × ^{nRL}P_{nRP} Where a = larger of n |

## Examples

The number of words that can be formed with the letters of the word Equation such that the even spaces are occupied by vowels

Number of letters

= 8 {E, Q, U, A, T, I, O, N}

⇒ n_{L} = 8

Number of places

= 8

⇒ n_{P} = 8

Number of specified letters

⇒ Number of Vowels

= 5

{E, U, A, I, O}

⇒ n_{SL} = 5

Number of specified places

⇒ Number of Even places

= 4

{X, _, X, _, X, _, X, _}

⇒ n_{SP} = 4

After filling the even places with vowels

Number of places filled

= 4

Smaller of n_{SL} (5) and n_{SP} (4)

⇒ n_{FP} = 4

Number of Places remaining to be filled

= Total Places − Places Filled

⇒ n_{RP} | = | n_{P} − n_{FP} |

= | 8 − 4 | |

= | 4 |

Number of Letters remaining to be used

= Total Letters − Places Filled

⇒ n_{RL} | = | n_{L} − n_{FP} |

= | 8 − 4 | |

= | 4 |

Number of words that can be formed with the letters of the word Equation such that the even spaces are occupied by vowels

= | ^{a}P_{b} × ^{nRL}P_{nRP} Where a is the larger of n |

= | ^{5}P_{4} × ^{4}P_{4} |

= | (5 × 4 × ... 4 terms) × 4! |

= | (5 × 4 × 3 × 2) × (4 × 3 × 2 × 1) |

= | 120 × 24 |

= | 2,880 |

## Using counting principles

**Total Event (E)**Filling the 8 places with the 8 letters

**1**^{st}sub-event (SE_{1})Filling the 4 even places with the 5 letters

Number of ways in which this event can be accomplished

= Number of ways in which 4 places can be filled with 5 letters Or = Number of permutations or arrangements with 5 letters taking 4 at a time ⇒ n _{SE1}= ^{5}P_{4}= 5 × 4 × 3 × 2 = 120 **2**^{nd}sub-event (SE_{2})Filling the remaining places with the remaining letters

Number of ways in which this event can be accomplished

= Number of ways in which the remaining 4 places can be filled with the remaining 4 letters Or = Number of permutations or arrangements with 4 letters taking 4 at a time ⇒ n _{SE2}= ^{4}P_{4}= 4! = 24

Number of words that can be formed with the letters of the word equation such that even places are occupied by vowels

⇒ n_{E} | = | n_{SE1} × n_{SE2} |

= | 120 × 24 | |

= | 2,880 |

The number of 8 letter words that can be formed with the letters of the word warehousing such that odd positions have only consonants.

Number of letters

= 11

{W, A, R, E, H, O, U, S, I, N, G}

⇒ n_{L} = 11

Number of places

Number of letters

= 8

⇒ n_{P} = 8

Number of specified letters

⇒ Number of Consonants

= 6

{W, R, H, S, N, G}

⇒ n_{SL} = 6

Number of specified places

⇒ Number of Odd Positions

= 4

{X, _, X, _, X, _, X, _}

⇒ n_{SP} = 4

After filling the odd places with consonants

Number of places filled

= 4

Smaller of n_{SL} (6) and n_{SP} (4)

⇒ n_{FP} = 4

Number of Places remaining to be filled

= Total Places − Places Filled

⇒ n_{RP} | = | n_{P} − n_{FP} |

= | 8 − 4 | |

= | 4 |

Number of Letters remaining to be used

= Total Letters − Places Filled

⇒ n_{RL} | = | n_{L} − n_{FP} |

= | 11 − 4 | |

= | 7 |

Number of 8 letter words that can be formed with the letters of the word warehousing such that the odd positions are filled with consonants

= | ^{a}P_{b} × ^{nRL}P_{nRP} Where a is the larger of n |

= | ^{6}P_{4} × ^{7}P_{4} |

= | (6 × 5 × ... 4 terms) × (7 × 6 × ... 4 terms) |

= | (6 × 5 × 4 × 3) × (7 × 6 × 5 × 4) |

= | 360 × 840 |

## Using counting principles

**Total Event (E)**Filling the 8 places with the 11 letters

**1**^{st}sub-event (SE_{1})Filling the 6 consonants in the 4 odd positions

Number of ways in which this can be accomplished

= Number of ways in which 4 positions can be filled with 6 letters Or = Number of permutations or arrangements of 6 items taking 4 at a time ⇒ n _{SE1}= ^{6}P_{4}= 6 × 5 × 4 × 3 = 360 **2**^{nd}sub-event (SE_{2})Filling the remaining 4 places with the remaining 7 letters

Number of ways in which this event can be accomplished

= Number of ways in which 4 places can be filled with the 7 letters Or = Number of permutations or arrangements with 7 items taking 4 at a time ⇒ n _{SE2}= ^{7}P_{4}= 7 × 6 × 5 × 4 = 840

The number of words that can be formed with the letters of the word warehousing such that odd positions are filled with consonants

⇒ n_{E} | = | n_{SE1} × n_{SE2} |

= | 360 × 840 |

# Two or more letters grouped (stay together)

n_{L} : Number of letters with which the words are to be formed

n_{P} : Number of places to fill the letters i.e. the number of letters in the word to be formed

n_{P} ≤ n_{L}

The maximum length of the word that can be formed is limited to the number of letters available.

n_{GL1} : Number of letters in the first group

n_{GL2} : Number of letters in the second group

n_{G} : Number of groups of letters

n_{GL} | : | Number of letters considering the letters to be grouped as a unit |

= | Total number of letters − Number of letters grouped + Number of groups of letters Remove all the letters that are grouped and add a letter for each group. |

⇒ n_{GL} | = | (n_{L} − $\sum _{i=1}^{{n}_{G}}{n}_{{G}_{{L}_{i}}}$) + n_{G} |

_{P}lettered words that can be formed using n

_{L}letters such that n

_{GL1}stay as a group, n

_{GL2}stay as another group, ... .

**Total Event (E)**Arranging the n

_{L}letters in n_{P}spaces

Assume that the total event is divided into n_{G} + 1 sub-events.

**1**^{st}sub-event (SE_{1})Arranging the n

_{GL}letters in as many places,Number of ways in which this event can be accomplished

= Number of ways in which n _{GL}places can be filled with as many lettersOr = Number of permutations or arrangements with n _{GL}items taking all at a time⇒ n _{SE1}= ^{nGL}P_{nGL}= n _{GL}!**2**^{nd}sub-event (SE_{2})Arranging the letters in the first group among themselves.

Number of ways in which this event can be accomplished

= Number of ways in which n _{GL1}places can be filled with as many lettersOr = Number of permutations or arrangements with n _{GL1}items taking all at a time⇒ n _{SE2}= ^{nGL1}P_{nGL1}= n _{GL1}!**3**^{rd}sub-event (SE_{3})Arranging the letters in the second group among themselves.

Number of ways in which this event can be accomplished

= Number of ways in which n _{GL2}places can be filled with as many lettersOr = Number of permutations or arrangements with n _{GL2}items taking all at a time⇒ n _{SE3}= ^{nGL2}P_{nGL2}= n _{GL2}!- ...
- ...

The number of words that can be formed

⇒ n_{E} | = | n_{SE1} × n_{SE2} × n_{SE3}× ... |

= | n_{GL}! × n_{GL1}! × n_{GL2}! × ... |

## Examples

The number of words that can be formed with the letters of the word Victory such that all the vowels come together

Number of letters

= 7 {V, I, C, T, O, R, Y}

⇒ n = 7

Number of Vowels

= 2 {I, O}

⇒ Number of letters in the group = 2

⇒ g = 2

Number of letters considering the vowels as a unit

= 6 {V, (I,O), C, T, R, Y}

⇒ n_{G} | = | 6 |

Or | = | (n − g) + 1 |

= | (7 − 2) + 1 | |

= | 6 |

Number of words that can be formed with the letters of the word Victory such that all the vowels come together

= | n_{G}! × g! |

= | 6! × 2! |

= | 1,440 |

## Using counting principles

**Total Event (E)**Filling the 7 places with the 7 letters

**1**^{st}sub-event (SE_{1})Arranging the letters taking the vowels as a unit in as many places

Number of ways in which this event can be accomplished

= Number of ways in which 6 places can be filled with as many letters Or = Number of permutations or arrangements with 6 letters taking all at a time ⇒ n _{SE1}= ^{6}P_{6}= 6! = 720 **2**^{nd}sub-event (SE_{2})Inter arranging the two vowels among themselves

Number of ways in which this event can be accomplished

= Number of ways in which 2 places containing vowels can be filled with 2 vowels Or = Number of permutations or arrangements with 2 letters taking all at a time ⇒ n _{SE2}= 2! = 2

Number of words that can be formed with the letters of the word victory such that all the vowels are together

⇒ n_{E} | = | n_{SE1} × n_{SE2} |

= | 720 × 2 | |

= | 1,440 |

The number of words that can be formed with the letters of the word Daughters such that all the vowels come together and DGHT stay together.

Number of letters

= 9 {D, A, U, G, H, T, E, R, S}

⇒ n = 9

Number of Vowels (first group)

= 3 {A, U, E}

⇒ g_{1} = 3

Number of letters in the second group

= 4 {D, G, H, T}

⇒ g_{2} = 4

Number of groups

⇒ N = 2

Number of letters considering the each of the letters grouped as a unit

= 4 {(A,U,E), (D,G,H,T), R, S}

⇒ n_{G} | = | 4 |

Or | = | [n − (g_{1} + g_{2})] + N |

= | [9 − (3 + 4)] + 2 | |

= | 9 − 7 + 2 | |

= | 4 |

Number of words that can be formed with the letters of the word Daughters such that all the vowels come together and DGHT stay together

= | n_{G}! × g_{1}! × g_{2}! |

= | 4! × 3! × 4! |

= | 24 × 6 × 24 |

= | 3,456 |

## Using counting principles

**Total Event (E)**Filling the 9 places with the 9 letters

**1**^{st}sub-event (SE_{1})Arranging the letters taking the vowels as a unit and DGHT as a unit in as many places

Number of ways in which this event can be accomplished

= Number of ways in which 4 places can be filled with as many letters Or = Number of permutations or arrangements with 4 letters taking all at a time ⇒ n _{SE1}= ^{4}P_{4}= 4! = 24 **2**^{nd}sub-event (SE_{2})Inter arranging the vowels among themselves

Number of ways in which this event can be accomplished

= Number of ways in which 3 places can be filled with as many letters Or = Number of permutations or arrangements with 3 letters taking all at a time ⇒ n _{SE2}= ^{3}P_{3}= 3! = 6 **3**^{rd}sub-event (SE_{3})Inter arranging the three letters DGHT among themselves

Number of ways in which this event can be accomplished

= Number of ways in which 4 places can be filled with as many letters Or = Number of permutations or arrangements with 4 letters taking all at a time ⇒ n _{SE3}= ^{4}P_{4}= 4! = 24

The number of words that can be formed with the letters of the word daughters such that all the vowels are together and the letters DGHT are together

⇒ n_{E} | = | n_{SE1} × n_{SE2} × n_{SE3} |

= | 24 × 6 × 24 | |

= | 3,456 |

# No two letters to come together

n_{L} : Number of letters with which the words are to be formed

n_{P} : Number of places to fill the letters i.e. the number of letters in the word to be formed

n_{P} ≤ n_{L}

The maximum length of the word that can be formed is limited to the number of letters available.

n_{DL} : Number of letters to stay divided/separate

n_{OL} | : | Number of other letters |

= | Total number of letters − Number of letters to stay divided/separate | |

= | (n_{L} − n_{DL}) |

### Places to arrange letters to stay separate

Using

- one letter we can keep two letters separate.
OL - two letters we can keep three letters separate
OL OL - ...
- n letters we can keep n + 1 letters separate

Thus in finding the number of places available to place the letters to stay separate, consider an arrangement of other letters with places on either side

OL_{1} | OL_{2} | OL_{3} | OL_{4} |

The places on either side of the other letters are the places where the letters to be divided/separated can appear to ensure that they do not come together.

n_{DP} | : | Number of places available to place the letters to say divided/separate |

= | Number of other letters + 1 | |

= | n_{OL} + 1 |

n_{DP} ≥ n_{DL}

To be able to keep n_{DL} letters separate we need at least as many spaces (n_{DP}) to fill them up.

If n_{DP} < n_{DL}, then it would not be possible to ensure that the n_{DL} letters stay separate.

#### n

_{DP}< n_{DL}**Eg**: Arrange the letters of the word UTOPIA such that no two vowels come together.In the word UTOPIANumber of letters

= 6

{U, T, O, P, I, A}

⇒ n

_{L}= 6Number of Letters to stay separate

⇒ Number of Vowels

= 4

{U, O, I, A}

⇒ n

_{DL}= 4Number of other letters

= Total number of letters − Number of letters to stay separate

⇒ n _{OL}= n _{L}− n_{DL}= 6 − 4 = 2 Number of places to place the letters to stay separate

= Number of other letters + 1

Since n⇒ n _{DP}= n _{OL}+ 1= 4 + 1 = 5 _{DP}< n_{DL}, it would not be possible to arrange the letters in such a way that the vowels do not come together.U T O P I A There are only two other letters, T and P. They can separate a maximum of 3 vowels. The fourth vowel would have to come beside another vowel.

#### n

When n_{DP}= n_{DL}_{DP}= n_{DL}, the specified letters would stay separate only if the word starts as well as ends with one of the letters to stay separate.**Eg**: Arrange the letters of the word Anxious such that no two vowels come together.In the word FortuneNumber of letters

= 7

{A, N, X, I, O, U, S}

⇒ n

_{L}= 7Number of Letters to stay separate

⇒ Number of Vowels

= 4

{A, I, O, U}

⇒ n

_{DL}= 4Number of other letters

= Total number of letters − Number of letters to stay separate

⇒ n _{OL}= n _{L}− n_{DL}= 7 − 4 = 3 Number of places to place the letters to stay separate

= Number of other letters + 1

Since n⇒ n _{DP}= n _{OL}+ 1= 3 + 1 = 4 _{DP}≥ n_{DL}, it would be possible to arrange the letters in such a way that the vowels do not come together.A N I X O S U We can neither start nor end the word with one of the other letters.

#### n

When n_{OL}= n_{DL}or n_{DP}= n_{DL}+ 1_{OL}= n_{DL}, ensuring that the letters to stay separate are arranged in the places specified for them would result in the other letters also staying separate.**Eg**: Arrange the letters of the word NATURE such that no two vowels come together.In the word NATURENumber of letters

= 6

{N, A, T, U, R, E}

⇒ n

_{L}= 6Number of Letters to stay separate

⇒ Number of Vowels

= 3

{A, U, E}

⇒ n

_{DL}= 3Number of other letters

= Total number of letters − Number of letters to stay separate

⇒ n _{OL}= n _{L}− n_{DL}= 6 − 3 = 3 Number of places to place the letters to stay separate

= Number of other letters + 1

Since n⇒ n _{DP}= n _{OL}+ 1= 2 + 1 = 3 _{DP}≥ n_{DL}, it would be possible to arrange the letters in such a way that the vowels do not come together.A N U R E T N U R E T A The letters would stay separated whether the word starts with a letter from the letters to stay separate or other letters.

**Total Event (E)**Arranging the n

_{L}letters in n_{P}spaces

Assume that the total event is divided into two sub-events.

**1**^{st}sub-event (SE_{1})Arranging the n

_{OL}letters in n_{OL}places.Number of ways in which this event can be accomplished

= Number of ways in which n _{OL}places can be filled with as many lettersOr = Number of permutations or arrangements with n _{OL}items taking all at a time⇒ n _{SE1}= ^{nOL}P_{nOL}= n _{OL}!**2**^{nd}sub-event (SE_{2})Arranging the n

_{DL}letters in the n_{DP}places that would ensure their staying separate,Number of ways in which this event can be accomplished

= Number of ways in which n _{DP}places can be filled with n_{DL}lettersOr = Number of permutations or arrangements with n _{DP}items taking n_{DL}at a time⇒ n _{SE2}= ^{nDP}P_{nDL}

The number of words that can be formed

⇒ n_{E} | = | n_{SE1} × n_{SE2} |

= | n_{OL}! × ^{nDP}P_{nDL} |

## Examples

The number of words that can be formed with the letters of the word DIALOGUE such that no two consonants come together

Number of letters

= 8

{D, I, A, L, O, G, U, E}

⇒ n_{L} = 8

Number of Letters to stay separate

⇒ Number of Consonants

= 3

{D, L, G}

⇒ n_{DL} = 3

Number of other letters

= Total number of letters − Number of letters to stay separate

⇒ n_{OL} | = | n_{L} − n_{DL} |

= | 8 − 3 | |

= | 5 |

Number of places to place the letters to stay separate

= Number of other letters + 1

⇒ n_{DP} | = | n_{OL} + 1 |

= | 5 + 1 | |

= | 6 |

_{DP}≥ n

_{DL}, it would be possible to arrange the letters in such a way that the vowels do not come together.

Number of words that can be formed with the letters of the word DIALOGUE such that no two consonants come together

= n_{OL}! × ^{nDP}P_{nDL}

= 5! × ^{(5 + 1)}P_{3}

= (5 × 4 × 3 × 2 × 1) × ^{6}P_{3}

= 120 × (6 × 5 × ... 3 terms)

= 120 × (6 × 5 × 4)

= 120 × 120

= 14,400

### Using Counting Principles

**Total Event (E)**Arranging the 8 letters in 8 spaces

**1**^{st}sub-event (SE_{1})Arranging the 5 other letters in 5 places.

Number of ways in which this event can be accomplished

= Number of ways in which 5 places can be filled with as many letters Or = Number of permutations or arrangements with 5 items taking all at a time ⇒ n _{SE1}= ^{5}P_{5}= 5! = 5 × 4 × 3 × 2 × 1 = 120 **2**^{nd}sub-event (SE_{2})Arranging the 3 letters in the 5 + 1 places around the other letters that would ensure their staying separate,

Number of ways in which this event can be accomplished

= Number of ways in which 6 places can be filled with 3 letters Or = Number of permutations or arrangements with 6 items taking 3 at a time ⇒ n _{SE2}= ^{6}P_{3}= 6 × 5 × ... 3 times = 6 × 5 × 4 = 120

The number of words that can be formed

⇒ n_{E} | = | n_{SE1} × n_{SE2} |

= | 120 × 120 | |

= | 14,400 |