# Identify the experiment, define the event, find the number of favorable choices in arranging all the Letters of a Word where all the letters are different

# Problem 4

- 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
- 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.
- The letters of the word RIGHTEOUSLY are arranged in a row at random. Find the probability that the even places are filled with consonants.
- 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
- The letters of the word TROPICAL are arranged in a row at random. Find the probability that no two vowels may come together.

# Solution 1

Number of Letters/Characters

= 5

{A, B, E, L, T}

⇒ n_{L} = 5

Experiment :

Forming a word using the 5 letters

Let

**A** : the event of forming the word TABLE

## For Event A

Number of Favorable Choices

= Number of ways in which the 5 letters can be arranged to form the word TABLE

= Number of ways in which T can be arranged in the first place × Number of ways in which A can be arranged in the second place × ...

To form the word TABLE, the letters T, A, B, L, E have to be fixed in the respective positions. Each letter has only one place where it can be filled.

= ^{1}P_{1} × ^{1}P_{1} × ...

= 1 × 1 × ...

= 1

⇒ m_{A} = 1

# Solution 2

Number of Letters/Characters

= 9

{S, L, A, U, G, H, T, E, R}

⇒ n_{L} = 9

Number of places to be filled in forming the required words

⇒ Number of letters in the word to be formed

= 9

⇒ n_{P} = 9

Experiment :

Forming a word using the 9 letters

Let

**B** : the event of the words formed having vowels in odd places

## For Event B

Number of specified places

⇒ Number of odd places in the word to be formed

= 5

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

⇒ n_{SP} = 5

Number of specified letters

⇒ Number of Vowels

= 3

{A, U, E}

⇒ n_{SL} = 3

After filling the even places with vowels

Number of places filled

= 3

Smaller of n_{SL} (3) and n_{SP} (5)

⇒ n_{FP} = 3

Number of Places remaining to be filled

= Total Places − Places Filled

⇒ n_{RP} | = | n_{P} − n_{FP} |

= | 9 − 3 | |

= | 6 |

Number of Letters remaining to be used

= Total Letters − Places Filled

⇒ n_{RL} | = | n_{L} − n_{FP} |

= | 9 − 3 | |

= | 6 |

Number of Favorable Choices

= Number of ways in which words can be formed with the letters of the word slaughter such that vowels occupy odd places

⇒ m_{B} | = | ^{a}P_{b} × ^{nRL}P_{nRP} Where a is the larger of n Number of ways in which the 3 vowels can be arranged in the 5 odd places × Number of ways in which the remaining 6 letters can be arranged in the remaining 6 places |

= | ^{5}P_{3} × ^{6}P_{6} | |

= | (5 × 4 × ... 3 terms) × 6! | |

= | (5 × 4 × 3) × (6 × 5 × 4 × 3 × 2 × 1) | |

= | 60 × 720 | |

= | 43,200 |

# Solution 3

Number of Letters/Characters

= 11

{R, I, G, H, T, E, O, U, S, L, Y}

⇒ n_{L} = 9

Number of places to be filled in forming the required words

⇒ Number of letters in the word to be formed

= 11

⇒ n_{P} = 11

Experiment :

Forming a word using the 11 letters

Let

**H** : the event of the words formed having even places filled with consonants

## For Event H

Number of specified places

⇒ Number of even places in the word to be formed

= 5

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

⇒ n_{SP} = 5

Number of specified letters

⇒ Number of Consonants

= 7

{R, G, H, T, S, L, Y}

⇒ n_{SL} = 7

After filling the even places with vowels

Number of places filled

= 3

Smaller of n_{SL} (7) and n_{SP} (5)

⇒ n_{FP} = 5

Number of Places remaining to be filled

= Total Places − Places Filled

⇒ n_{RP} | = | n_{P} − n_{FP} |

= | 11 − 5 | |

= | 6 |

Number of Letters remaining to be used

= Total Letters − Places Filled

⇒ n_{RL} | = | n_{L} − n_{FP} |

= | 11 − 5 | |

= | 6 |

Number of Favorable Choices

= Number of ways in which words can be formed with the letters of the word righteously such that even places are filled with consonants

⇒ m_{H} | = | ^{a}P_{b} × ^{nRL}P_{nRP} Where a is the larger of n Number of ways in which the 5 even places can be filled with the 7 consonants × Number of ways in which the remaining 6 letters can be arranged in the remaining 6 places |

= | ^{7}P_{5} × ^{6}P_{6} | |

= | (7 × 6 × ... 5 terms) × 6! | |

= | (7 × 6 × 5 × 4 × 3) × (6 × 5 × 4 × 3 × 2 × 1) | |

= | 2,520 × 720 | |

= | 18,14,400 |

# Solution 4

Number of Letters/Characters

= 8

{P, E, N, T, A, G, O, N}

⇒ n_{L} = 8

Experiment :

Forming a word using the 8 letters

Let

**T** : the event of forming the words with exactly two letters between E and A

## For Event T

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

- The First and the Fourth places
{_, X, X, _, X, X, X, X}

- The Second and the Fifth places
{X, _, X, X, _, X, X, X}

- The Third and the Sixth places
{X, X, _, X, X, _, X, X}

- The Fourth and the Seventh places
{X, X, X, _, X, X, _, X}

- The Fifth and the Eighth places
{X, X, X, X, _, X, X, _ }

Number of Favorable Choices

= Number of ways in which words can be formed with the letters of the word righteously with exactly two letters between E and A

⇒ m_{T} | = | Number of ways in which one of the 5 alternatives for placing E and A can be chosen × Number of ways in the letters E and A can be placed in the two places × Number of ways in which the remaining 6 letters can be arranged in the remaining 6 places |

= | ^{5}C_{1} × ^{2}P_{2} × ^{6}P_{6} | |

= | 5 × 2! × 6! | |

= | 5 × (2 × 1) × (6 × 5 × 4 × 3 × 2 × 1) | |

= | 5 × 2 × 720 | |

= | 7,200 |

### Alternative

We can find the number of ways in which each alternative for event T can be accomplished and use the fundamental counting theorem of addition to find the total number of favorable choices for event T.### Event T_{EA1}

Number of Favorable Choices = Number of ways in which words can be formed with the letters of the word righteously with E and A occupying the first and third places

⇒ m_{TEA1} | = | Number of ways in the letters E and A can be placed in the two places × Number of ways in which the remaining 6 letters can be arranged in the remaining 6 places |

= | ^{2}P_{2} × ^{6}P_{6} | |

= | 2! × 6! | |

= | (2 × 1) × (6 × 5 × 4 × 3 × 2 × 1) | |

= | 2 × 720 | |

= | 1,440 |

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

⇒ m_{TEA1} = m_{TEA1} = m_{TEA1} = m_{TEA1} = m_{TEA1} = 1,440

Total Number of Favorable Choices for event T

= Sum of the number of ways in which each alternative event can be accomplished.

⇒ m_{T} | = | Σm_{TEAi} |

= | m_{TEA1} + m_{TEA1} + m_{TEA1} + m_{TEA1} + m_{TEA1} | |

= | 1,440 + 1,440 + 1,440 + 1,440 + 1,440 | |

= | 7,200 |

# Solution 5

Number of Letters/Characters

= 8

{T, R, O, P, I, C, A, L}

⇒ n_{L} = 8

Number of places to be filled in forming the required words

⇒ Number of letters in the word to be formed

= 8

⇒ n_{P} = 8

Experiment :

Forming a word using the 8 letters

Let

**L** : the event of the words formed with no two vowels together

## For Event L

Number of Letters to stay separate

⇒ Number of Vowels

= 3

{O, I, A}

⇒ n_{DL} = 3

Number of other letters

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

⇒ n_{OL} | = | n_{L} − n_{DL} |

= | 8 − 3 | |

= | 5 |

n_{DL} − 1 | = | 3 − 1 |

= | 2 |

_{OL}≥ n

_{DL}− 1, it would be possible to arrange the letters in such a way that the vowels do not come together.

Number of Favorable Choices

= Number of ways in which words can be formed with the letters of the word tropical with no two vowels coming together

= Number of ways in which the 5 other letters can be arranged in 5 places to create places around them for arranging the letters to stay separate × Number of ways in the 3 vowels to stay separate can be arranged in the 6 (5 + 1) places created around the 5 other letters

⇒ m_{L} | = | ^{5}P_{5} × ^{6}P_{3} |

= | 5! × (6 × 5 × ... 3 terms) | |

= | (5 × 4 × 3 × 2 × 1) × (6 × 5 × 4) | |

= | 120 × 120 | |

= | 14,400 |