Probability : Problem Solving : Arranging Letters of a Word (letters are all different)

Permutations/Arrangements

The number of permutations or arrangements that can be made with "n" different things taking "r" at a time is given by

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

Permutations/Arrangements taking only some of the letters of a word (letters are all different)

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

Permutations/Arrangements taking all the letters of a word (letters are all different)

The number of words that can be formed using the letters of a "n" letter word taking all the letters ("r" = "n") at a time and where the letters are all different is given by
"The number of permutations of "n" letters" i.e. nPn (= n!)
nPn =
n!
(n − n)!
 ⇒ nPn =
n!
(0)!
 ⇒ nPn =
n!
1
 ⇒ nPn = n!

• Explanation

Let us assume that there are "n" places into which "n" letters of a word are arranged. Since they are to be re-arranged, 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.

         
1st 2nd ... ... nth

The total event of placing the "n" letters in the "n" places an be divided into the sub-events of placing each letter starting from the first.

  • Total Event (E) = Arranging the "n" letters in "n" spaces
  • 1st sub-event (E1) = Arranging a letter in the First place

    The 1st place can be filled first with any one of the available "n" letters

    This can be done in "n" ways ⇒ nE1 = n.

  • 2nd sub-event (E2) = Arranging a letter in the Second place

    The 2nd place can be filled with any one of the (n − 1) letters remaining after filling the 1st place

    This can be done "n − 1" ways ⇒ nE2 = n − 1.

  • 3rd sub-event (E3) = Arranging a letter in the third place

    The 3rd place can be filled with any one of the (n − 2) letters remaining after filling the 1st and the 2nd places

    This can be done "n − 2" ways ⇒ nE3 = n − 2.

  • . . .
  • . . .
  • nth sub-event (En) = Arranging a letter in the nth place

    The nth place can be filled with the 1 letter (n − (n − 1) remaining after filling the first (n − 1) places

    This can be done is "1" way ⇒ nEn = 1.

Therefore, the no. of ways in which the "n" letters can be filled in the "n" places is given by

nE = nE1 × nE2 × nE3 × .... × nEn
= (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
      = nPr where n = 7 and r = 7 (Or) n! where n = 7
      = 7P7 or 7!
      = 7!

Words formed by fixing one or more letters in specific places

• Fixing one Letter

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 sub-events.
  • Total Event (E) » Arranging the "n" letters in "n" spaces
  • 1st sub-event (E1) » Arranging the fixed letter in the required place

    This can be done in one way
  • 2nd sub-event (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 nE = nE1 × nE2
= 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
  • 1st sub-event (E1) » Filling the middle place with "T"

    No. of ways in which this can be accomplished ⇒ nE1 = 1P1 (Or) 1!
    = 1

  • 2nd sub-event (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 ⇒ nE2 = 6P6 (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"

⇒ nE = nE1 × nE2
= 1 × 720
= 720

• Fixing 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
    • 1st sub-event (E1) » Arranging the first letter to be fixed in the required place

      This can be done in one way
    • 2nd sub-event (E2) » Arranging the second letter to be fixed in the required place

      This can be done in one way
    • 3rd sub-event (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 nE = nE1 × nE2 × nE3
    = 1 × 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
  • 1st sub-event (E1) » Filling the first place with "T"

    No. of ways in which this can be accomplished ⇒ nE1 = 1P1 (Or) 1!
    = 1

  • 2nd sub-event (E2) » Filling the last place with "Y"

    No. of ways in which this can be accomplished ⇒ nE2 = 1P1 (Or) 1!
    = 1

  • 3rd sub-event (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 ⇒ nE3 = 6P6 (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"

⇒ nE = nE1 × nE2 × nE3
= 1 × 1 × 720
= 720

Words formed by fixing a set of letters in a set of places

• Fixing a Set of 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 sub-events.
  • Total Event (E) » Arranging the "n" letters in "n" spaces
  • 1st sub-event (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)

    • When a > b, The no. of possibilities = a P b

      ["The no. of permutations of "a" letters taking "b" places at a time]

    • Wheen b > a, The no. of possibilities = b P a

      ["The no. of permutations of "b" places taking "a" letters at a time]

    • When a = b, The no. of possibilities = aPa i.e. a! (Or) bPb i.e. b!
  • 2nd sub-event (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
    ⇒ nc = 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 nE = nE1 × nE2

  • When a > b,     nE = aPb × (n − b)!
  • When b > a,     nE = bPa × (n − a)!
  • When a = b,     nE = a! (or b!) × (n − a)! (or (n − b!))

» Illustration

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
  • 1st sub-event (E1) » Filling the 4 even places with the 5 letters

    No. of ways in which this can be accomplished ⇒ nE1 = 5P4
    = 5 × 4 × 3 × 2
    = 120

  • 2nd sub-event (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 ⇒ nE2 = 4P4 (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.

⇒ nE = nE1 × nE2
= 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
  • 1st sub-event (E1) » Filling the 3 letters in the 4 odd positions

    No. of ways in which this can be accomplished ⇒ nE1 = 4P3
    = 4 × 3 × 2
    = 24

  • 2nd sub-event (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 ⇒ nE2 = 4P4 (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

⇒ nE = nE1 × nE2
= 24 × 24
= 576

• Fixing more than one Set of Letters in that many Sets of Places

Where there are two or more sets specified, arranging letters within each set is to be treated as a sub event.

Words with two or more letters grouped (stay together)

• One Group

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 sub-events.
  • Total Event (E) » Arranging the "n" letters in "n" spaces
  • 1st sub-event (E1) » Arranging ((n − a) + 1) letters in as many places by considering 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)!

  • 2nd sub-event (E2) » Arranging the letters which stay together among themselves.

    The number of ways this can be done ⇒ aPa (Or) a!

The no. of words that can be formed ⇒ nE = nE1 × nE2
= ((n − a) + 1)! × a!

» Illustration

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
  • 1st sub-event (E1) » Filling the 6 places with the 6 letters by taking the vowels as group

    No. of ways in which this can be accomplished ⇒ nE1 = 6P6 (Or) 6!
    = 6 × 5 × 4 × 3 × 2 × 1
    = 720

  • 2nd sub-event (E2) » Inter arranging the two vowels among themselves

    No. of ways in which this can be accomplished ⇒ nE2 = 2P2 (Or) 2!
    = 2 × 1
    = 2

The no. of words that can be formed with the letters of the word victory such that all the vowels are together

⇒ nE = nE1 × nE2
= 720 × 2
= 1,440

• Two or More Groups

Where there are two or more groups, each group is to be treated as a separate entity

» Illustration

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 grouped letters (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
  • 1st sub-event (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 ⇒ nE1 = 4P4 (Or) 4!
    = 4 × 3 × 2 × 1
    = 24
  • 2nd sub-event (E2) » Inter arranging the three vowels among themselves

    No. of ways in which this can be accomplished ⇒ nE2 = 3P3 (Or) 3!
    = 3 × 2 × 1
    = 6

  • 3rd sub-event (E3) » Inter arranging the three letters "DGH" among themselves

    No. of ways in which this can be accomplished ⇒ nE3 = 3P3 (Or) 3!
    = 3 × 2 × 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

⇒ nE = nE1 × nE2 × nE3
= 24 × 6 × 6
= 864

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