# Permutations/Arrangements

The number of permutations with n different things taking r at a time is given by nPr, where
nPr =
 n! (n − r)!
= (n)! × (n − 1)! × (n − 2)! × .... r times

# Permutations/Arrangements in making words from letters

nL : Number of letters with which the words are to be formed

nP : Number of places to fill the letters i.e. the number of letters in the word to be formed

nP ≤ nL

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

Number of nP lettered words that can be formed using nL letters where all the letters are different

⇒ Number of words that can be formed using nL letters taking nP letters at a time

⇒ Number of permutations of nL items taking nP at a time

nLPnP =
 nL! (nL − nP)!
= (nL) × (nL − 1) × (nL − 2) × .... nP times

## Explanation

Forming a nP letter word with nL letters can be assumed as the act of arranging nL letters into nP places.
 1st 2nd ... ... nPth
• Total Event (E)

Arranging the nL letters in nP places

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

• 1st sub-event (SE1)

Placing a letter in the First place

The 1st place can be filled with any one of the available nL letters

⇒ SE1 can be accomplished in nL ways

⇒ nSE1 = nL

• 2nd sub-event (SE2)

Placing a letter in the Second place

The 2nd place can be filled with any one of the remaining nL − 1 letters

⇒ SE2 can be accomplished in nL − 1 ways

⇒ nSE2 = nL − 1

• 3rd sub-event (SE3)

Placing a letter in the Third place

The 3rd place can be filled with any one of the remaining nL − 2 letters

⇒ SE3 can be accomplished in nL − 2 ways

⇒ nSE3 = nL − 2

• ...
• ...
• nPth sub-event (SEnP)

Placing a letter in the nPth place

The nPth place can be filled with any one of the nL − (nP − 1) letters remaining after filling the first nP − 1 places

⇒ SE3 can be accomplished in nL − (nP − 1) ways

⇒ nSEnp = nL − nP + 1

By the fundamental counting principle of multiplication,

Number of ways in which the nL letters can be filled in the nP places

 ⇒ nE = nSE1 × nSE2 × nSE3 × .... × nSEp = (nL) × (nL − 1) × (nL − 2) × ... × (nL − nP + 1) = (nL) × (nL − 1) × (nL − 2) × ... nP times ≡ n × (n − 1) × (n − 2) × ... × (n − r + 1) = nLPnP ≡ nPr

## Example

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

In the word subdermatoglyphic

Number of letters

= 17

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

⇒ nL = 17

Number of letters in the word to be formed

⇒ Number of places to be filled in forming the word

= 5

⇒ nP = 5

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

= Number of words that can be formed using nL letters taking nP letters at a time

= Number of permutations of nL items taking nP at a time

= nLPnP

= 17P5

= 17 × 16 × 15 × 14 × 13

## Where all letters are used

Where all letters are used in forming the words,

nL = nP

Number of ways in which the nL letters can be filled in the nP places

 ⇒ nE = nLPnP = nLPnL = nL! Or = nPPnP = nP!

## Example

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

In the word Algebra

Number of letters

= 7

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

⇒ nL = 7

Number of letters in the word to be formed

⇒ Number of places to be filled in forming the word

= 7

⇒ nP = 7

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

= Number of words that can be formed using nL letters taking nP letters at a time

= Number of permutations of nL items taking nP at a time

= nLPnP

= nL!

= 7!

= 4,090

# Fixing letters (each in its own place)

nL : Number of letters with which the words are to be formed

nP : Number of places to fill the letters i.e. the number of letters in the word to be formed

nP ≤ nL

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

nFL : Number of letters to be fixed each in its own place

After fixing nFL letters each in its own place

Number of Letters remaining

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

⇒ nRL = nL − nFL

Number of Places remaining to be filled

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

⇒ nRP = nP − nFL

Number of nP letter words that can be formed using nL letters by fixing nFL letters each in its own place is given by

nRLPnRP

## Explanation

• Total Event (E)

Arranging the nL letters in nP spaces

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

• 1st sub-event (SE1)

Arranging the nFL letters each in its own place

Number 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

⇒ nSE1 = 1

• 2nd sub-event (SE2)

Filling the remaining nRP places with the remaining nRL letters

Number of ways in which this event can be accomplished

 = Number of ways in which nRP places can be filled with the nRL letters Or = Number of permutations or arrangements of nRL items taking nRP items at a time
 ⇒ nSE2 = nRLPnRP

The number of ways in which the nL letters can be filled in the nP places with nFL letters fixed each in its own place

 ⇒ nE = nSE1 × nSE2 = 1 × nRLPnRP = nRLPnRP

### 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

In the word incomputably

Number of letters in the word

= 12

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

nL = 12

Number of places to be filled in forming the words

= 7

nP = 7

Number of letters fixed each in its own place

nFL = 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

 ⇒ nRL = nL − nFL = 12 − 1 = 11

Number of places remaining to be filled

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

 ⇒ nRP = nP − nFL = 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

= nRLPnRP

= 11P6

= 11 × 10 × 9 × 8 × 7 × 6

### using fundamental principle

• Total Event (E)

Filling the 7 places with the 12 letters

• 1st sub-event (SE1)

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
 ⇒ nSE1 = 1P1 = 1! = 1
• 2nd sub-event (SE2)

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
 ⇒ nSE2 = 11P6

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

 ⇒ nE = nSE1 × nSE2 = 1 × 11P6 = 11P6 = 11 × 10 × 9 × 8 × 7 × 6

## Where all letters are used

Where all letters are used in forming the words,

nL = nP

⇒ nL − nFL = nP − nFL

⇒ nRL = nRP

Number of nP lettered words that can be formed with the nL letters, fixing nFP letters each in its own place

 = nRLPnRP = nRLPnRL = nRL! Or = nRPPnRP = nRP!

## 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

In the word thursday,

Number of letters

= 8

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

nL = 8

Number of places to be filled in forming the words

= 8

nP = 8

Number of letters fixed each in its own place

nFL = 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

 ⇒ nRL = nL − nFL = 8 − 2 = 6

Number of places remaining to be filled

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

 ⇒ nRP = nP − nFL = 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

= nRLPnRP

= nRLPnRL Or nRPPnRP

= nRL! or nRP!

= 6!

= 720

### using fundamental principle

• Total Event (E)

Filling the 8 places with the 8 letters

• 1st sub-event (SE1)

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

⇒ nSE1 = 1

• 2nd sub-event (SE2)

Filling the remaining places with the remaining letters

Number of ways in which this event can be accomplished

 = Number of ways in which nRP remaining places can be filled with the nRL remaining letters Or = Number of permutations or arrangements with nRP items taking all at a time
 ⇒ nSE2 = nRLPnRP = 6P6 = 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

 ⇒ nE = nSE1 × nSE2 = 1 × 720 = 720

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

nL : Number of letters with which the words are to be formed

nP : Number of places to fill the letters i.e. the number of letters in the word to be formed

nP ≤ nL

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

nSL : Number of letters specified to be filled in the specified places

nSP : Number of places specified to be filled with the specified letters

nFP : Number of places filled

If nSL ≠ nSP,

Number of places filled would be the lesser of nSL and nSP.

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.

⇒ nFP = smaller of nSL and nSP

Where nSL = nSP,

nFP = nSP = nSL

After fixing nFL letters each in its own place

Number of Letters remaining

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

⇒ nRL = nL − nFL

Number of Places remaining to be filled

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

⇒ nRP = nP − nFL

nRP ≤ nRL

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

Number of nP letter words that can be formed using nL letters by fixing nFL letters each in its own place is given by

nRLPnRP

## Explanation

• Total Event (E)

Arranging the nL letters in nL spaces

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

• 1st sub-event (SE1)

Arranging the nSL specified letters in the nSP specified places

Number of ways in which this event can be accomplished

 = Number of ways in which the nSP specified places can be filled with the nSL specified letters Or = Number of permutations or arrangements with greater of nSP and nSL items taking the lesser of them at a time
 ⇒ nSE1 = aPb Number of permutations of a items taking b at a time Where a = larger of nSL and nSP and b the other If nSL = nSP, a = b = nSL = nSP
• 2nd sub-event (SE2)

Arranging the remaining letters in the remaining places.

Number of ways in which this event can be accomplished

 = Number of ways in which nRP places can be filled with the nRL letters Or = Number of permutations or arrangements of nRL items taking nRP at a time
 ⇒ nSE2 = nRLPnRP

The number of words that can be formed

 ⇒ nE = nSE1 × nSE2 = aPb × nRLPnRP Where a = larger of nSL and nSP and b the other

## 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

In the word Equation

Number of letters

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

⇒ nL = 8

In the word to be formed

Number of places

= 8

⇒ nP = 8

Number of specified letters

⇒ Number of Vowels

= 5

{E, U, A, I, O}

⇒ nSL = 5

Number of specified places

⇒ Number of Even places

= 4

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

⇒ nSP = 4

After filling the even places with vowels

Number of places filled

= 4

Smaller of nSL (5) and nSP (4)

⇒ nFP = 4

Number of Places remaining to be filled

= Total Places − Places Filled

 ⇒ nRP = nP − nFP = 8 − 4 = 4

Number of Letters remaining to be used

= Total Letters − Places Filled

 ⇒ nRL = nL − nFP = 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

 = aPb × nRLPnRP Where a is the larger of nSL (5) and nSP (4) and b the other = 5P4 × 4P4 = (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

• 1st sub-event (SE1)

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
 ⇒ nSE1 = 5P4 = 5 × 4 × 3 × 2 = 120
• 2nd sub-event (SE2)

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
 ⇒ nSE2 = 4P4 = 4! = 24

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

 ⇒ nE = nSE1 × nSE2 = 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.

In the word hypnotizable

Number of letters

= 11

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

⇒ nL = 11

In the words to be formed

Number of places

Number of letters

= 8

⇒ nP = 8

Number of specified letters

⇒ Number of Consonants

= 6

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

⇒ nSL = 6

Number of specified places

⇒ Number of Odd Positions

= 4

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

⇒ nSP = 4

After filling the odd places with consonants

Number of places filled

= 4

Smaller of nSL (6) and nSP (4)

⇒ nFP = 4

Number of Places remaining to be filled

= Total Places − Places Filled

 ⇒ nRP = nP − nFP = 8 − 4 = 4

Number of Letters remaining to be used

= Total Letters − Places Filled

 ⇒ nRL = nL − nFP = 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

 = aPb × nRLPnRP Where a is the larger of nSL (6) and nSP (4) and b the other = 6P4 × 7P4 = (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

• 1st sub-event (SE1)

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
 ⇒ nSE1 = 6P4 = 6 × 5 × 4 × 3 = 360
• 2nd sub-event (SE2)

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
 ⇒ nSE2 = 7P4 = 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

 ⇒ nE = nSE1 × nSE2 = 360 × 840

# Two or more letters grouped (stay together)

nL : Number of letters with which the words are to be formed

nP : Number of places to fill the letters i.e. the number of letters in the word to be formed

nP ≤ nL

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

nGL1 : Number of letters in the first group

nGL2 : Number of letters in the second group

nG : Number of groups of letters

 nGL : 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.
 ⇒ nGL = (nL − $\sum _{i=1}^{{n}_{G}}{n}_{{G}_{{L}_{i}}}$) + nG
Number of nP lettered words that can be formed using nL letters such that nGL1 stay as a group, nGL2 stay as another group, ... .
• Total Event (E)

Arranging the nL letters in nP spaces

Assume that the total event is divided into nG + 1 sub-events.

• 1st sub-event (SE1)

Arranging the nGL letters in as many places,

Number of ways in which this event can be accomplished

 = Number of ways in which nGL places can be filled with as many letters Or = Number of permutations or arrangements with nGL items taking all at a time
 ⇒ nSE1 = nGLPnGL = nGL!
• 2nd sub-event (SE2)

Arranging the letters in the first group among themselves.

Number of ways in which this event can be accomplished

 = Number of ways in which nGL1 places can be filled with as many letters Or = Number of permutations or arrangements with nGL1 items taking all at a time
 ⇒ nSE2 = nGL1PnGL1 = nGL1!
• 3rd sub-event (SE3)

Arranging the letters in the second group among themselves.

Number of ways in which this event can be accomplished

 = Number of ways in which nGL2 places can be filled with as many letters Or = Number of permutations or arrangements with nGL2 items taking all at a time
 ⇒ nSE3 = nGL2PnGL2 = nGL2!
• ...
• ...

The number of words that can be formed

 ⇒ nE = nSE1 × nSE2 × nSE3× ... = nGL! × nGL1! × nGL2! × ...

## Examples

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

In the word Victory

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}

 ⇒ nG = 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

 = nG! × g! = 6! × 2! = 1,440

## Using counting principles

• Total Event (E)

Filling the 7 places with the 7 letters

• 1st sub-event (SE1)

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
 ⇒ nSE1 = 6P6 = 6! = 720
• 2nd sub-event (SE2)

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
 ⇒ nSE2 = 2! = 2

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

 ⇒ nE = nSE1 × nSE2 = 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.

In the word Daughter

Number of letters

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

⇒ n = 9

Number of Vowels (first group)

= 3 {A, U, E}

⇒ g1 = 3

Number of letters in the second group

= 4 {D, G, H, T}

⇒ g2 = 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}

 ⇒ nG = 4 Or = [n − (g1 + g2)] + 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

 = nG! × g1! × g2! = 4! × 3! × 4! = 24 × 6 × 24 = 3,456

## Using counting principles

• Total Event (E)

Filling the 9 places with the 9 letters

• 1st sub-event (SE1)

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
 ⇒ nSE1 = 4P4 = 4! = 24
• 2nd sub-event (SE2)

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
 ⇒ nSE2 = 3P3 = 3! = 6
• 3rd sub-event (SE3)

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
 ⇒ nSE3 = 4P4 = 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

 ⇒ nE = nSE1 × nSE2 × nSE3 = 24 × 6 × 24 = 3,456

# No two letters to come together

nL : Number of letters with which the words are to be formed

nP : Number of places to fill the letters i.e. the number of letters in the word to be formed

nP ≤ nL

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

nDL : Number of letters to stay divided/separate

 nOL : Number of other letters = Total number of letters − Number of letters to stay divided/separate = (nL − nDL)

### 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

 OL1 OL2 OL3 OL4

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.

 nDP : Number of places available to place the letters to say divided/separate = Number of other letters + 1 = nOL + 1

nDP ≥ nDL

To be able to keep nDL letters separate we need at least as many spaces (nDP) to fill them up.

If nDP < nDL, then it would not be possible to ensure that the nDL letters stay separate.

• #### nDP < nDL

Eg : Arrange the letters of the word UTOPIA such that no two vowels come together.

In the word UTOPIA

Number of letters

= 6

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

⇒ nL = 6

Number of Letters to stay separate

⇒ Number of Vowels

= 4

{U, O, I, A}

⇒ nDL = 4

Number of other letters

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

 ⇒ nOL = nL − nDL = 6 − 4 = 2

Number of places to place the letters to stay separate

= Number of other letters + 1

 ⇒ nDP = nOL + 1 = 4 + 1 = 5
Since nDP < nDL, 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.

• #### nDP = nDL

When nDP = nDL, 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 Fortune

Number of letters

= 7

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

⇒ nL = 7

Number of Letters to stay separate

⇒ Number of Vowels

= 4

{A, I, O, U}

⇒ nDL = 4

Number of other letters

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

 ⇒ nOL = nL − nDL = 7 − 4 = 3

Number of places to place the letters to stay separate

= Number of other letters + 1

 ⇒ nDP = nOL + 1 = 3 + 1 = 4
Since nDP ≥ nDL, 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.

• #### nOL = nDL or nDP = nDL + 1

When nOL = nDL, 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 NATURE

Number of letters

= 6

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

⇒ nL = 6

Number of Letters to stay separate

⇒ Number of Vowels

= 3

{A, U, E}

⇒ nDL = 3

Number of other letters

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

 ⇒ nOL = nL − nDL = 6 − 3 = 3

Number of places to place the letters to stay separate

= Number of other letters + 1

 ⇒ nDP = nOL + 1 = 2 + 1 = 3
Since nDP ≥ nDL, 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 nL letters in nP spaces

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

• 1st sub-event (SE1)

Arranging the nOL letters in nOL places.

Number of ways in which this event can be accomplished

 = Number of ways in which nOL places can be filled with as many letters Or = Number of permutations or arrangements with nOL items taking all at a time
 ⇒ nSE1 = nOLPnOL = nOL!
• 2nd sub-event (SE2)

Arranging the nDL letters in the nDP places that would ensure their staying separate,

Number of ways in which this event can be accomplished

 = Number of ways in which nDP places can be filled with nDL letters Or = Number of permutations or arrangements with nDP items taking nDL at a time
 ⇒ nSE2 = nDPPnDL

The number of words that can be formed

 ⇒ nE = nSE1 × nSE2 = nOL! × nDPPnDL

## Examples

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

In the word DIALOGUE

Number of letters

= 8

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

⇒ nL = 8

Number of Letters to stay separate

⇒ Number of Consonants

= 3

{D, L, G}

⇒ nDL = 3

Number of other letters

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

 ⇒ nOL = nL − nDL = 8 − 3 = 5

Number of places to place the letters to stay separate

= Number of other letters + 1

 ⇒ nDP = nOL + 1 = 5 + 1 = 6
Since nDP ≥ nDL, 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

= nOL! × nDPPnDL

= 5! × (5 + 1)P3

= (5 × 4 × 3 × 2 × 1) × 6P3

= 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

• 1st sub-event (SE1)

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
 ⇒ nSE1 = 5P5 = 5! = 5 × 4 × 3 × 2 × 1 = 120
• 2nd sub-event (SE2)

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
 ⇒ nSE2 = 6P3 = 6 × 5 × ... 3 times = 6 × 5 × 4 = 120

The number of words that can be formed

 ⇒ nE = nSE1 × nSE2 = 120 × 120 = 14,400