# Arranging all the Letters of a Word where all the letters are different

No. Problems & Solutions [Click Hide/Show to display the solutions below the question]
04.
Define the Event and identify the number of favourable choices in the following cases
 a) The letters of the word SLAUGHTER are arranged in a row at random. Find the probability that the vowels may be in the odd places. b) The letters of the word "RIGHTEOUSLY" are arranged in a row at random. Find the probability that the even places are filled with consonants. c) The letters of the word “PENTAGON” are arranged in a row at random. Find the probability that there are exactly two letters between E and A is d) Each of the letters A, B, E, L, T are written on a separate card. If all the cards are arranged in a row in all possible ways, the probability of the forming the word TABLE is

 Solution (a)
 ⇒ nL Number of Letters/Characters in the word "SLAUGHTER" = 9 {S, L, A, U, G, H, T, E, R} = 9

In the experiment of testing for the number of words that can be formed using the letters of the word "SLAUGHTER"

 ⇒ n Total No. of Possible Choices = Number of words that can be formed using the 9 letters of the word "SLAUGHTER" = 9P9 (Or) 9!
Let "A" be the event of the word formed with all the vowels occupying odd places

# For Event "A"

No. of vowels (a) = 3 {A, U, E}
No. of other letters (Consonants) = 6 {S, L, G, H, T, R}
No. of odd places = 5 {_, X, _, X, _, X, _, X, _}

 ⇒ mA Number of Favourable/Favorable Choices = The no. of words that can be formed using the letters of the word "SLAUGHTER" such that the vowels occupy odd places. = (No. of ways in which the 3 vowels can be arranged in the 5 odd places) × (No. of ways in which the remaining 6 letters can be arranged       in the remaining 6 places) = 5P3 × 6P6 = 5 × 4 × 3 × 6! = 60 × 720 = 43,200

 Solution (b)
 ⇒ nL Number of Letters/Characters in the word "RIGHTEOUSLY" = 11{R, I, G, H, T, E, O, U, S, L, Y} = 11

In the experiment of testing for the number of words that can be formed using the letters of the word "RIGHTEOUSLY"

 ⇒ n Total No. of Possible Choices = Number of words that can be formed using the 9 letters of the word "RIGHTEOUSLY" = 11P11 (Or) 11!
Let "H" be the event of forming the words such that the consonants occupy even places

# For Event "H"

No. of Consonants = 7 {R, G, H, T, S, L, Y}
No. of other letters (Vowels) = 4 {I, E, O, U}
No. of even places = 5 {X, _, X, _, X, _, X, _, X, _, X}

 ⇒ mH Number of Favourable/Favorable Choices = The no. of words that can be formed using the letters of the word "RIGHTEOUSLY" such that the even places are filled with consonants = (No. of ways in which the 5 even places can be filled with the 7       consonants) × (No. of ways in which the remaining 6 places can be filled with the       remaining 6 letters) = 7P5 × 6P6 = 7 × 6 × 5 × 4 × 3 × 6! = 2,520 × 720 = 18,14,400

 Solution (c)
 ⇒ nL Number of Letters/Characters in the word "PENTAGON" = 8 {P, E, N, T, A, G, O, N} = 8

In the experiment of testing for the number of words that can be formed using the letters of the word "PENTAGON"

 ⇒ n Total No. of Possible Choices = Number of words that can be formed using the 8 letters of the word "PENTAGON" = 8P8 (Or) 8!
Let "T" be the event of forming the words such that there are exactly two letters between E and A

# For Event "T"

Event "T" can be accomplished in five alternative ways, with "E" and "A" occupying

1. The First and the Fourth places {_, X, X, _, X, X, X, X} → (Ta)
2. The Second and the Fifth places {X, _, X, X, _, X, X, X} → (Tb)
3. The Third and the Sixth places {X, X, _, X, X, _, X, X} → (Tc)
4. The Fourth and the Seventh places {X, X, X, _, X, X, _, X} → (Td)
5. The Fifth and the Eighth places {X, X, X, X, _, X, X, _ } → (Te)

## • 1st Alternative (Ta)

 ⇒ mTa Number of Favourable/Favorable Choices = The no. of words that can be formed using the letters of the word "PENTAGON" filling E and A in the first and third places. = (No. of ways in which the 6 letters other than "E" and "A" can       be filled in the "6" places) × (No. of ways in which "E" and "A" can be inter arranged in the       two places designated to them) = 6P6 × 2P2 = 6! × 2! = 720 × 2 = 1,440

Similarly, each of the alternative events can be accomplished in 1,440 ways.

Thus, mTa = mTb = mTc = mTd = mTe = 1,440

 ⇒ mT Total Number of Favourable/Favorable Choices = Sum of the number of ways in which each alternative event     can be accomplished. = mTa + mTb + mTc + mTd + mTe = 1,440 + 1,440 + 1,440 + 1,440 + 1,440 = 7,200

 Solution (d)
 ⇒ nL Number of Letters/Characters given = 5 {A, B, E, L, T} = 5

In the experiment of testing for the number of words that can be formed using the given letters

 ⇒ n Total No. of Possible Choices = Number of words that can be formed using the 5 letters = 5P5 (Or) 5! = 120
Let "B" be the event of forming the words such that the word formed is TABLE

# For Event "B"

For forming the word "TABLE" each letter has to be fixed in a particular place.

 Number of Favourable/Favorable Choices = (No. of ways in which the Letter "T" can be filled in the First place) × (No. of ways in which the Letter "A" can be filled in the Second place) × (No. of ways in which the Letter "B" can be filled in the Third place) × (No. of ways in which the Letter "L" can be filled in the Fourth place) × (No. of ways in which the Letter "E" can be filled in the Fifth place) = 1P1 × 1P1 × 1P1 × 1P1 × 1P1 = 1 × 1 × 1 × 1 × 1 = 1

 Author Credit : The Edifier