## Problem Solving : Arranging Letters of a Word (isograms - letters are all different) |

## 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 ^{n}P_{r}

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

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

## What are isograms? |

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.

- Words in which no letter is used more than once.
- subdermatoglyphic
- hydropneumatics
- troublemaking

- 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. ^{L}P_{P}### Illustration

Explanation Show

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

1^{st} | 2^{nd} | ... | ... | n^{th} |

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
- E
_{1}(1^{st}sub-event) : Arranging a letter in the First placeThe 1

^{st}place can be filled first with any one of the available "L" lettersThis can be done in "L" ways ⇒ n

_{E1}= L. - E
_{2}(2^{nd}sub-event) : Arranging a letter in the Second placeThe 2

^{nd}place can be filled with any one of the (L - 1) letters remaining after filling the 1^{st}placeThis can be done in "L - 1" ways ⇒ n

_{E2}= L - 1. - E
_{3}(3^{nd}sub-event) : Arranging a letter in the Third placeThe 3

^{rd}place can be filled with any one of the (L - 2) letters remaining after filling the 1^{st}and the 2^{nd}placesThis can be done in "L - 2" ways ⇒ n

_{E3}= L - 2. - . . .
- . . .
- E
_{n}(n^{th}sub-event) : Arranging a letter in the P^{th}place.The P

^{th}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 ⇒ n

_{En}= (L - P + 1).

By the fundamental counting theorem of multiplication,

n_{E} | = | n_{E1} × n_{E2} × n_{E3} × .... × n_{En} |
---|---|---|

= | (L) × (L - 1) × (L - 2) × ... × (L - P + 1) | |

= | ^{L}P_{P} |

- 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. = ^{L}P_{P}= (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 - 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. = ^{L}P_{P}= ^{7}P_{7}= 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### E

_{1}: 1^{st}sub-eventArranging the L_{(F)}letters to be fixed in that many P_{(F)}placesThis can be done in one way since each letter is to be fixed in its own place

### E

_{2}: 2^{nd}sub-eventArranging the remaining L_{(R)}letters in the remaining P_{(R)}placesThis 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) | ||

⇒ n_{E} | = | n_{E1} × n_{E2} | By the fundamental counting theorem of multiplication. |
---|---|---|---|

= | 1 × ^{L(R)} P _{P(R)} | ||

= | ^{L(R)} P _{P(R)} |

- 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 EventFilling the 7 places with the 7 letters [P with L]

**E**: 1_{1}^{st}sub-eventFilling the fixed places with the fixed letters i.e the middle place with 'T'.

**E**: 2_{2}^{nd}sub-eventFilling 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

⇒ n _{E}= n _{E1}× n_{E2}= 1 × ^{L(R)}P_{P(R)}= ^{6}P_{6}= 1 × (6 × 5 × 4 × 3 × 2 × 1) = 720 - 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 EventFilling the 5 places with the 10 letters [P with L]

**E**: 1_{1}^{st}sub-eventFilling the fixed places with the fixed letters i.e the first place with 'M'.

**E**: 2_{2}^{nd}sub-eventFilling 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

⇒ n _{E}= n _{E1}× n_{E2}= 1 × ^{L(R)}P_{P(R)}= ^{9}P_{4}= 9 × 8 × 7 × 6 = 3,024

- 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 EventFilling the 8 places with the 8 letters [P with L]

**E**: 1_{1}^{st}sub-eventFilling the fixed places with the fixed letters i.e the first place with 'T' and the last place with 'Y'.

**E**: 2_{2}^{nd}sub-eventFilling 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

⇒ n _{E}= n _{E1}× n_{E2}= 1 × ^{L(R)}P_{P(R)}= ^{6}P_{6}= (6 × 5 × 4 × 3 × 2 × 1) = 720 - 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 EventFilling the 6 places with the 9 letters [P with L]

**E**: 1_{1}^{st}sub-eventFilling the fixed places with the fixed letters i.e the first place with 'S' and the last place with 'e'.

**E**: 2_{2}^{nd}sub-eventFilling 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

⇒ n _{E}= n _{E1}× n_{E2}= 1 × ^{L(R)}P_{P(R)}= ^{7}P_{4}= (7 × 6 × 5 × 4) = 840

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

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)} |

Arranging the "L" letters in "P" places

Arranging the L_{(S)} specified letters in the P_{(S)} specified places.

Numer of ways in which the 1^{st} 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)} | ||

⇒ n_{E1} | = | ^{P(S)} P _{L(S)} | |
---|---|---|---|

(Or) | = | Number of permutations of L_{(S)} entities taking P_{(S)} at a time Where L_{(S)} > P_{(S)} | |

⇒ n_{E1} | = | ^{L(S)} P _{P(S)} |

Arranging the remaining L_{(R)} letters in the remaining P_{(R)} places.

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

= | Number of ways in which the remaining L_{(R)} letters can be arranged in the remaining P_{(R)} places. | |

⇒ n_{E2} | = | ^{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). | |

⇒ n_{E} | = | n_{E1} × n_{E2} |
---|---|---|

= | ^{L(S)} P _{P(S)} × ^{L(R)} P _{P(R)} |

- 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)}= 4After 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 EventFilling the 8 places with the 8 letters [P with L].

**E**: 1_{1}^{st}sub-eventFilling the 4 even places with 5 vowels [P

_{(S)}with L_{(S)}]**E**: 2_{2}^{nd}sub-eventFilling 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

⇒ n _{E}= n _{E1}× n_{E2}= ^{L(S)}P_{P(S)}×^{L(R)}P_{P(R)}Since L_{(S)}> P_{(S)}= ^{5}P_{4}×^{4}P_{4}= (5 × 4 × 3 × 2) × (4 × 3 × 2 × 1) = 120 × 24 = 2,880 - 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)}= 3After 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 EventFilling the 7 places with the 7 letters. [P with L]

**E**: 1_{1}^{st}sub-eventFilling the 4 odd places with the 3 consonants [P

_{(S)}with L_{(S)}]**E**: 2_{2}^{nd}sub-eventFilling 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

⇒ n _{E}= n _{E1}× n_{E2}= ^{P(S)}P_{L(S)}×^{L(R)}P_{P(R)}Since P_{(S)}> L_{(S)}= ^{4}P_{3}×^{4}P_{4}= (4 × 3 × 2) × (4 × 3 × 2 × 1) = 24 × 24 = 576 - 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)}= 4After 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 EventFilling the 7 places with the 12 letters [P with L]

**E**: 1_{1}^{st}sub-eventFilling the 4 odd places with the 8 consonants [P

_{(S)}with L_{(S)}]**E**: 2_{2}^{nd}sub-eventFilling 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

⇒ n _{E}= n _{E1}× n_{E2}= ^{L(S)}P_{P(S)}×^{L(R)}P_{P(R)}Since L_{(S)}> P_{(S)}= ^{8}P_{4}×^{8}P_{3}= (8 × 7 × 6 × 5) × (8 × 7 × 6 ) = 1,680 × 336 = 5,64,480 - 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)}= 5After 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 EventFilling the 9 places with the 11 letters [P with L]

**E**: 1_{1}^{st}sub-eventFilling the first six places with 5 vowels [P

_{(S)}with L_{(S)}]**E**: 2_{2}^{nd}sub-eventFilling 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

⇒ n _{E}= n _{E1}× n_{E2}= ^{L(S)}P_{P(S)}×^{L(R)}P_{P(R)}Since L_{(S)}> P_{(S)}= ^{6}P_{5}×^{6}P_{4}= (6 × 5 × 4 × 3 × 2) × (6 × 5 × 4 × 3) = 720 × 360 = 2,59,200

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

**E**: Total EventArranging the L letters in P places

**E**: 1_{1}^{st}sub-eventArranging the first set of specified letters L

_{(S1)}in the specified places P_{(S1)}.**E**: 2_{2}^{nd}sub-eventArranging the second set of specified letters L

_{(S2)}in the specified places P_{(S2)}.- ...
- ...
**E**: n_{n}^{th}sub-eventArranging 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) × ... | |

⇒ n_{E} | = | n_{E1} × n_{E2} × ... × n_{En} |
---|---|---|

= | 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.

### • One Group (Illustrations)

### • Two or More Groups (Illustrations)

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### E

_{1}: 1^{st}sub-eventArranging L_{(G)}letters in as many places.### E

_{2}: 2^{nd}sub-eventInter-arranging the first set of grouped letters (L_{(G1)}) among themselves.### E

_{3}: 3^{rd}sub-eventInter-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) × ... | |

⇒ n_{E} | = | n_{E1} × n_{E2} × n_{E2} × ... |
---|---|---|

= | ^{L(G)} P _{L(G)} × ^{L(G1)} P _{L(G1)} × ^{L(G2)} P _{L(G2)} × ... | |

= | L_{(G)}! × L_{(G1)}! × L_{(G2)}! × ... |

- 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 EventFilling the 7 places with the 7 letters [P with L]

**E**: 1_{1}^{st}sub-eventArranging L

_{(G)}letters in as many places.**E**: 2_{2}^{nd}sub-eventInter 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

⇒ n _{E}= n _{E1}× n_{E2}= ^{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

- 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 EventFilling the 8 places with the 8 letters [P with L]

**E**: 1_{1}^{st}sub-eventArranging L

_{(G)}letters in as many places.**E**: 2_{2}^{nd}sub-eventInter arranging the L

_{(G1)}grouped letters within the first group.**E**: 3_{3}^{rd}sub-eventInter arranging the L

_{(G2)}grouped letters within the second group.

Number of words that can be formed

⇒ n _{E}= n _{E1}× n_{E2}× n_{E3}= ^{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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