Problem Solving : Arranging Letters of a Word (isograms - 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

nPr =
n!
(n - r)!
(Or) = (n) × (n - 1) × (n - 2) × .... r times.

What are isograms?

Isogram

Meaning

A line drawn on a map connecting points having the same numerical value of some variable.

It is used with the sense in recreational linguistics.

Examples

  1. Words in which no letter is used more than once.
    • subdermatoglyphic
    • hydropneumatics
    • troublemaking
  2. Words in which each letter appears the same number of times, not necessarily just once.
    • uncopyrightable [Each letter appears 1 time],
    • intestines [Each letter appears 2 times],
    • scintillescent [Each letter appears 2 times],
    • sestettes [Each letter appears 3 times],

Words formed with all/some of the letters of a word (letters are all different)

The number of words that can be formed using "P" letters of a "L" letter word where the letters are all different is given by "the number of permutations of "L" letters taking "P" at a time" i.e. LPP

Explanation Show
There are "P" places into which "L" letters of the word are arranged.

1st 2nd ... ... nth

The total event of placing the "L" letters in the "P" places can be divided into the independent sub-events of placing each letter starting from the first.

  • E (Total Event) : Arranging "L" letters in "P" places
  • E1 (1st sub-event) : Arranging a letter in the First place

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

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

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

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

    This can be done in "L - 1" ways ⇒ nE2 = L - 1.

  • E3 (3nd sub-event) : Arranging a letter in the Third place

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

    This can be done in "L - 2" ways ⇒ nE3 = L - 2.

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

    The Pth place can be filled with the letter (L - (P - 1) remaining after filling the first (P - 1) places.

    This can be done in (L - (P - 1)) ways i.e. (L - P + 1) ways ⇒ nEn = (L - P + 1).

By the fundamental counting theorem of multiplication,

nE = nE1 × nE2 × nE3 × .... × nEn
= (L) × (L - 1) × (L - 2) × ... × (L - P + 1)
= LPP

Illustration

  1. Number of two letter words that can be formed using the letters of the word MARCH, without repeating any letter.

    Solution Show

    In the Word given

    L = Number of letters
    = 5 {M, A, R, C, H}

    In the word to be formed,

    P = Number of letters (Or) Places to be filled for forming the word
    = 2

    Number of 2 letter words that can be formed using the letters of the word 'MARCH'

    = Number of permutations of 5 (L) different things taking 2 (P) at a time.
    = LPP
    = (L) × (L - 1 ) × ... P times
    = (5) × (5 - 1 )
    = 5 × 4
    = 20

    Check
    MA, MR, MC, MH, AR, AC, AH, RC, RH, CH
    AA, RR, CM, HM, RA, CA, HA, CR, HR, HC

  2. The no. of words that can be formed with the letters of the word "Almonds"

    Solution Show

    In the Word given

    L = Number of letters
    = 7 {A, l, m, o, n, d, s}

    In the word to be formed,

    P = Number of letters (Or) Places to be filled for forming the word
    = 7

    Number of 7 letter words that can be formed using the letters of the word 'Almonds'

    = Number of permutations of 7 (L) different things taking 7 (P) at a time.
    = LPP
    = 7P7
    = 7!
    = 5,040

Words formed by fixing one or more letters in specific places

To Find the number of words that can be formed using 'P' letters of a "L" letter word and fixing L(F) letters each in its own place, assume that the total event is divided into two independent sub-events.

Where,

In the Word given

L = Number of letters

In the word to be formed,

P = Number of letters (Or) Places to be filled for forming the word
L(F) = Number of letters fixed each in its own place
P(F) = Number of places to be filled with fixed letters

L(F) = P(F).

In the word to be formed, after fixing the letters in the specified places

P(R) = Number of places remaining to be filled
= Total Places − Places with fixed letters
= P − P(F)
L(R) = Number of letters remaining for filling the remaining places
= Total Letters in the word given − Letters fixed in fixed places
= L − L(F)

  • E : Total Event

    Arranging the "L" letters in "P" spaces
  • E1 : 1st sub-event

    Arranging the L(F) letters to be fixed in that many P(F) places

    This can be done in one way since each letter is to be fixed in its own place

  • E2 : 2nd sub-event

    Arranging the remaining L(R) letters in the remaining P(R) places

    This can be done in L(R) P P(R) ways

The no. of words that can be formed using "L" letters by fixing "L(F)" letters each in its own place

= (No. of ways in which the letters to be fixed in the specified places can be arranged in those places) × (No. of ways in which the remaining P(R) places can be filled with the remaining L(R) letters)
⇒ nE = nE1 × nE2
By the fundamental counting theorem of multiplication.
= 1 × L(R) P P(R)
= L(R) P P(R)

• Fixing one Letter (Illustrations)

  1. The no. of words that can be formed with all the letters of the word "Tuesday" such that "T" occupies the middle place

    Solution Show

    In the Word given

    L = Number of letters
    = 7 {T, u, e, s, d, a, y}

    In the word to be formed,

    P = Number of letters (Or) Places to be filled for forming the word
    = 7
    L(F) = Number of letters fixed each in its own place
    = 1
    P(F) = Number of places to be filled with fixed letters
    = 1

    L(F) = P(F) = 1.

    In the word to be formed, after fixing the letters in the specified places

    P(R) = Number of places remaining to be filled
    = Total Places − Places with fixed letters
    = P − P(F)
    = 7 − 1
    = 6
    L(R) = Number of letters remaining for filling the remaining places
    = Total Letters in the word given − Letters fixed in fixed places
    = L − L(F)
    = 7 − 1
    = 6

    • E : Total Event

      Filling the 7 places with the 7 letters [P with L]

    • E1 : 1st sub-event

      Filling the fixed places with the fixed letters i.e the middle place with 'T'.

    • E2 : 2nd sub-event

      Filling the remaining 6 places with the remaining 6 letters [P(R) with L(R)]

    The number of words that can be formed with all the letters of the word 'Tuesday' such that 'T' occupies the middle place

    ⇒ nE = nE1 × nE2
    = 1 × L(R) P P(R)
    = 6P6
    = 1 × (6 × 5 × 4 × 3 × 2 × 1)
    = 720

  2. The no. of 5 letter words that can be formed with the letters of the word "Magnitudes" such that "M" occupies the first place

    Solution Show

    In the Word given

    L = Number of letters
    = 10 {M, a, g, n, i, t, u, d, e, s}

    In the word to be formed,

    P = Number of letters (Or) Places to be filled for forming the word
    = 5
    L(F) = Number of letters fixed each in its own place
    = 1
    P(F) = Number of places to be filled with fixed letters
    = 1

    L(F) = P(F) = 1.

    In the word to be formed, after fixing the letters in the specified places

    P(R) = Number of places remaining to be filled
    = Total Places − Places with fixed letters
    = P − P(F)
    = 5 − 1
    = 4
    L(R) = Number of letters remaining for filling the remaining places
    = Total Letters in the word given − Letters fixed in fixed places
    = L − L(F)
    = 10 − 1
    = 9

    • E : Total Event

      Filling the 5 places with the 10 letters [P with L]

    • E1 : 1st sub-event

      Filling the fixed places with the fixed letters i.e the first place with 'M'.

    • E2 : 2nd sub-event

      Filling the remaining 4 places with the remaining 9 letters [P(R) with L(R)]

    The number of 5 letter words that can be formed with the letters of the word 'Magnitudes' such that 'M' occupies the first place

    ⇒ nE = nE1 × nE2
    = 1 × L(R) P P(R)
    = 9P4
    = 9 × 8 × 7 × 6
    = 3,024

• Fixing two or more Letters (each in its own place) [Illustrations]

  1. The no. of words that can be formed with all the letters of the word "Thursday" such that "T" occupies the first place and "Y" occupies the last place

    Solution Show

    In the Word given

    L = Number of letters
    = 8 {T, h, u, r, s, d, a, y}

    In the word to be formed,

    P = Number of letters (Or) Places to be filled for forming the word
    = 8
    L(F) = Number of letters fixed each in its own place
    = 2
    P(F) = Number of places to be filled with fixed letters
    = 2

    L(F) = P(F) = 2.

    In the word to be formed, after fixing the letters in the specified places

    P(R) = Number of places remaining to be filled
    = Total Places − Places with fixed letters
    = P − P(F)
    = 8 − 2
    = 6
    L(R) = Number of letters remaining for filling the remaining places
    = Total Letters in the word given − Letters fixed in fixed places
    = L − L(F)
    = 8 − 2
    = 6

    • E : Total Event

      Filling the 8 places with the 8 letters [P with L]

    • E1 : 1st sub-event

      Filling the fixed places with the fixed letters i.e the first place with 'T' and the last place with 'Y'.

    • E2 : 2nd sub-event

      Filling the remaining 6 places with the remaining 6 letters [P(R) with L(R)]

    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

    ⇒ nE = nE1 × nE2
    = 1 × L(R) P P(R)
    = 6P6
    = (6 × 5 × 4 × 3 × 2 × 1)
    = 720

  2. The number of six letter words that can be formed with the letters of the word "Symbolize" such that "S" occupies the first place and "e" occupies the last place

    Solution Show

    In the Word given

    L = Number of letters
    = 9 {S, y, m, b, o, l, i, z, e}

    In the word to be formed,

    P = Number of letters (Or) Places to be filled for forming the word
    = 6
    L(F) = Number of letters fixed each in its own place
    = 2
    P(F) = Number of places to be filled with fixed letters
    = 2

    L(F) = P(F) = 2.

    In the word to be formed, after fixing the letters in the specified places

    P(R) = Number of places remaining to be filled
    = Total Places − Places with fixed letters
    = P − P(F)
    = 6 − 2
    = 4
    L(R) = Number of letters remaining for filling the remaining places
    = Total Letters in the word given − Letters fixed in fixed places
    = L − L(F)
    = 9 − 2
    = 7

    • E : Total Event

      Filling the 6 places with the 9 letters [P with L]

    • E1 : 1st sub-event

      Filling the fixed places with the fixed letters i.e the first place with 'S' and the last place with 'e'.

    • E2 : 2nd sub-event

      Filling the remaining 4 places with the remaining 7 letters [P(R) with L(R)]

    The number of words that can be formed with the letters of the word 'Symbolize' such that 'S' occupies the first place and "e" occupies the last place

    ⇒ nE = nE1 × nE2
    = 1 × L(R) P P(R)
    = 7P4
    = (7 × 6 × 5 × 4)
    = 840

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

Fixing one set of Letters

To find the number of words that can be formed using "P" letters of a "L" letter word by arranging some of the letters in specified places, assume that the total event is divided into two sub-events.

Where

L = Number of letters in the given Word
P = Number of letters in the word to be formed
(Or) Places to be filled for forming the new word
L(S) = Number of specified letters for the word to be formed
P(S) = Number of specified places in the word to be formed

In filling the specified places with the specified letters

L(U) = Number of letters used up
P(U) = Number of places used up

L(U) = P(U) = Lower of L(S) and P(S).

Where

  • L(S)< P(S), as many places as there are letters can only be used up ⇒ L(U) = P(U) = L(S).
  • P(S)< L(S), as many letters as there are places can only be used up ⇒ L(U) = P(U) = P(S).

After fixing the letters in the specified places,

P(R) = Number of places remaining to be filled in the word to be formed
= Total Places in the word to be formed
− Places used up in filling the Specified Places with Specified Letters
= P − P(U)
L(R) = Number of letters remaining for filling the remaining places
= Total Letters in the word given
− Letters used up in filling the Specified Places with Specified Letters
= L − L(U)

E : Total Event

Arranging the "L" letters in "P" places

E1 : 1st sub-event

Arranging the L(S) specified letters in the P(S) specified places.

Numer of ways in which the 1st sub-event can be accomplished

= Number of ways in which P(S) places can be filled with L(S) letters
= Number of permutations of P(S) entities taking L(S) at a time Where P(S) > L(S)
⇒ nE1 = P(S) P L(S)
(Or) = Number of permutations of L(S) entities taking P(S) at a time Where L(S) > P(S)
⇒ nE1 = L(S) P P(S)

E2 : 2nd sub-event

Arranging the remaining L(R) letters in the remaining P(R) places.

Number of ways in which the 2nd sub-event can be accomplished

= Number of ways in which the remaining L(R) letters can be arranged in the remaining P(R) places.
⇒ nE2 = L(R) P P(R)

Number of words that can be formed

= (Number of ways in which the L(S) specified letters can be arranged in the P(S) specified places ) × (Number of ways in which the remaining L(R) letters can be arranged in the remaining P(R) places).
⇒ nE = nE1 × nE2
= L(S) P P(S) × L(R) P P(R)

Illustrations

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

    Solution Show

    L = Number of letters in the word Equation
    = 8 {E, q, u, a, t, i, o, n}

    P = Number of places in the word to be formed
    = 8

    L(S) = Number of specified letters for the word to be formed
    = Number of Vowels in the word Equation {E, u, a, i, o}
    = 5
    P(S) = Number of specified places in the word to be formed
    = Number of Even places in the word to be formed {X, _, X, _, X, _, X, _}
    = 4

    In filling the even places with vowels

    L(U) = Number of letters used up
    P(U) = Number of places used up

    Since P(S)< L(S),

    L(U) = P(U) = P(S) ⇒ L(U) = P(U) = 4

    After filling the even places with vowels

    P(R) = Number of places remaining to be filled
    = P − P(U)
    = 8 − 4
    = 4
    L(R) = Number of letters remaining
    = L − L(U)
    = 8 − 4
    = 4

    • E : Total Event

      Filling the 8 places with the 8 letters [P with L].

    • E1 : 1st sub-event

      Filling the 4 even places with 5 vowels [P(S) with L(S)]

    • E2 : 2nd sub-event

      Filling the remaining 4 places with the remaining 4 letters [P(R) with L(R)]

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

    ⇒ nE = nE1 × nE2
    = L(S) P P(S) × L(R) P P(R) Since L(S) > P(S)
    = 5P4 × 4P4
    = (5 × 4 × 3 × 2) × (4 × 3 × 2 × 1)
    = 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.

    Solution Show

    L = Number of letters in the word FAILURE
    = 7 {F, A, I, L, U, R, E}

    P = Number of places in the word to be formed
    = 7

    L(S) = Number of specified letters for the word to be formed
    = Number of Consonants in the word FAILURE {F, L, R}
    = 3
    P(S) = Number of specified places in the word to be formed
    = Number of Odd places in the word to be formed {_, X, _, X, _, X, _}
    = 4

    In filling the odd places with consonants

    L(U) = Number of letters used up
    P(U) = Number of places used up

    Since L(S)< P(S),

    L(U) = P(U) = L(S) ⇒ L(U) = P(U) = 3

    After filling the odd places with consonants

    P(R) = Number of places remaining to be filled
    = P − P(U)
    = 7 − 3
    = 4
    L(R) = Number of letters remaining
    = L − L(U)
    = 7 − 3
    = 4

    • E : Total Event

      Filling the 7 places with the 7 letters. [P with L]

    • E1 : 1st sub-event

      Filling the 4 odd places with the 3 consonants [P(S) with L(S)]

    • E2 : 2nd sub-event

      Filling the remaining 4 places with the remaining 4 letters [P(R) with L(R)]

    The number of words that can be formed with all the letters of the word 'FAILURE' such that the consonants occupying odd positions

    ⇒ nE = nE1 × nE2
    = P(S) P L(S) × L(R) P P(R) Since P(S) > L(S)
    = 4P3 × 4P4
    = (4 × 3 × 2) × (4 × 3 × 2 × 1)
    = 24 × 24
    = 576

  3. The number of seven letter words that can be formed with the letters of the word "considerably" such that the odd spaces are occupied by consonants.

    Solution Show

    L = Number of letters in the word considerably
    = 12 {c, o, n, s, i, d, e, r, a, b, l, y}

    P = Number of places in the word to be formed
    = 7

    L(S) = Number of specified letters for the word to be formed
    = Number of Consonants in the word considerably {C, N, S, D, R, B, L, Y}
    = 8
    P(S) = Number of specified places in the word to be formed
    = Number of Odd places in the word to be formed {_, X, _, X, _, X, _}
    = 4

    In filling the odd places with consonants

    L(U) = Number of letters used up
    P(U) = Number of places used up

    Since P(S)< L(S),

    L(U) = P(U) = P(S) ⇒ L(U) = P(U) = 4

    After filling the odd places with consonants,

    P(R) = Number of places remaining to be filled
    = Total Places − Lower of Specified Letters/Specified Places
    = P − P(U)
    = 7 − 4
    = 3
    L(R) = Number of letters remaining
    = Total Letters − Lower of Specified Letters/Specified Places
    = L − L(U)
    = 12 − 4
    = 8

    • E : Total Event

      Filling the 7 places with the 12 letters [P with L]

    • E1 : 1st sub-event

      Filling the 4 odd places with the 8 consonants [P(S) with L(S)]

    • E2 : 2nd sub-event

      Filling the remaining 3 places with the remaining 8 letters [P(R) with L(R)]

    The number of words that can be formed with the letters of the word 'considerably' such that the odd spaces are occupied by cononants

    ⇒ nE = nE1 × nE2
    = L(S) P P(S) × L(R) P P(R) Since L(S) > P(S)
    = 8P4 × 8P3
    = (8 × 7 × 6 × 5) × (8 × 7 × 6 )
    = 1,680 × 336
    = 5,64,480

  4. Nine letters of the word FACETIOUSLY are arranged at random. The number of words that can be formed with the vowels occupying the first six positions.

    Solution Show

    L = Number of letters in the word FACETIOUSLY
    = 11 {F, A, C, E, T, I, O, U, S, L, Y}

    P = Number of places in the word to be formed
    = 9

    L(S) = Number of specified letters for the word to be formed
    = Number of Vowels in the word FACETIOUSLY {A, E, I, O, U}
    = 5
    P(S) = Number of specified places in the word to be formed
    = First six places {__, __, _,__, __, _, X, X, X} in the word to be formed
    = 6

    In filling the first six positions with vowels

    L(U) = Number of letters used up
    P(U) = Number of places used up

    Since P(S)< L(S),

    L(U) = P(U) = P(S) ⇒ L(U) = P(U) = 5

    After filling the specified places with the specified letters,

    P(R) = Number of places remaining to be filled
    = P − P(U)
    = 9 − 5
    = 4
    L(R) = Number of letters remaining
    = L − L(U)
    = 11 − 5
    = 6

    • E : Total Event

      Filling the 9 places with the 11 letters [P with L]

    • E1 : 1st sub-event

      Filling the first six places with 5 vowels [P(S) with L(S)]

    • E2 : 2nd sub-event

      Filling the remaining 4 places with the remaining 6 letters [P(R) with L(R)]

    The number of words that can be formed with all the letters of the word 'FACETIOUSLY' such that the vowels occupy the first six positions

    ⇒ nE = nE1 × nE2
    = L(S) P P(S) × L(R) P P(R) Since L(S) > P(S)
    = 6P5 × 6P4
    = (6 × 5 × 4 × 3 × 2) × (6 × 5 × 4 × 3)
    = 720 × 360
    = 2,59,200

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.
  • E : Total Event

    Arranging the L letters in P places

  • E1 : 1st sub-event

    Arranging the first set of specified letters L(S1) in the specified places P(S1).

  • E2 : 2nd sub-event

    Arranging the second set of specified letters L(S2) in the specified places P(S2).

  • ...
  • ...
  • En : nth sub-event

    Arranging the remaining letters L(R) in the remaining spaces P(R).

The number of words that can be formed

= (Number of ways in which the first sub-event can be accomplished)
× (Number of ways in which the second sub-event can be accomplished)
× ...
⇒ nE = nE1 × nE2 × ... × nEn
= P( L(S1), P(S1) ) × P( L(S2), P(S2) ) × ... × P( L(R), P(R) )
P( L(S1), P(S1) ) if L(S1) > P(S1) and P( P(S1), L(S1) ) if P(S1) > L(S1)

Words with two or more letters grouped (stay together)

For finding the number of words that can be formed using "P" letters of a "L" letter word such that two or more letters stay together in one or more groups, assume the total event to be divided into number of letter groups + 1 sub events.

Where,

In the Word given

L = Number of letters

In the word to be formed,

P = Number of letters (Or) Places to be filled for forming the word
G = Number of letters groups
L(G1) = Number of letters in the first group.
L(G2) = Number of letters in the second group.
... =
L(G) = Total Number of letters, taking each letter group as a unit (single letter).
= L − (L(G1) + L(G2)) + ... ) + G.

  • E : Total Event

    Arranging the "L" letters in "P" places
  • E1 : 1st sub-event

    Arranging L(G) letters in as many places.
  • E2 : 2nd sub-event

    Inter-arranging the first set of grouped letters (L(G1)) among themselves.
  • E3 : 3rd sub-event

    Inter-arranging the second set of grouped letters (L(G2) among themselves.
  • ...

Number of words that can be formed

= (Number of ways in which the L(G) letters can be arranged in as many places)
× (Number of ways in which L(G1) letters can be inter arranged between themselves)
× (Number of ways in which L(G2) letters can be inter arranged between themselves)
× ...
⇒ nE = nE1 × nE2 × nE2 × ...
= L(G) P L(G) × L(G1) P L(G1) × L(G2) P L(G2) × ...
= L(G)! × L(G1)! × L(G2)! × ...

• One Group (Illustrations)

  1. The no. of words that can be formed with the letters of the word "Victory" such that all the vowels come together

    Solution Show

    In the Word given

    L = Number of letters
    = 7 {V, i, c, t, o, r, y}

    In the word to be formed,

    P = Number of letters (Or) Places to be filled for forming the word
    = 7
    G = Number of letters groups
    = 1
    L(G1) = Number of letters in the first group.
    = 2 {Vowels - I, O}
    L(G) = Total Number of letters, taking each letter group as a unit (single letter).
    = L − (L(G1)) + G.
    = 7 − (2) + 1.
    = 6

    • E : Total Event

      Filling the 7 places with the 7 letters [P with L]

    • E1 : 1st sub-event

      Arranging L(G) letters in as many places.

    • E2 : 2nd sub-event

      Inter arranging the L(G1) grouped letters within the group.

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

    ⇒ nE = nE1 × nE2
    = L(G) P L(G) × L(G1) P L(G1)
    (Or) = L(G)! × L(G1)!
    = 6! × 2!
    = (6 × 5 × 4 × 3 × 2 × 1) × (2 × 1)
    = 720 × 2
    = 1,440

• Two or More Groups (Illustrations)

  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

    Solution Show

    In the Word given

    L = Number of letters
    = 8 {D, a, u, g, h, t, e, r}

    In the word to be formed,

    P = Number of letters (Or) Places to be filled for forming the word
    = 8
    G = Number of letters groups
    = 2
    L(G1) = Number of letters in the first group.
    = 3 {Vowels - A, U, E}
    L(G2) = Number of letters in the second group.
    = 3 {D, G, H}
    L(G) = Total Number of letters, taking each letter group as a unit (single letter).
    = L − (L(G1) + L(G2)) + G.
    = 8 − (3 + 3) + 1.
    = 3

    • E : Total Event

      Filling the 8 places with the 8 letters [P with L]

    • E1 : 1st sub-event

      Arranging L(G) letters in as many places.

    • E2 : 2nd sub-event

      Inter arranging the L(G1) grouped letters within the first group.

    • E3 : 3rd sub-event

      Inter arranging the L(G2) grouped letters within the second group.

    Number of words that can be formed

    ⇒ nE = nE1 × nE2 × nE3
    = L(G) P L(G) × L(G1) P L(G1) × L(G2) P L(G2)
    = L(G)! × L(G1)! × L(G2)!
    = 4! × 3! × 3!
    = (4 × 3 × 2 × 1) × (3 × 2 × 1) × (3 × 2 × 1)
    = 24 × 6 × 6
    = 864

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