# Problem 4

Define the Event and identify the number of favourable choices in the following cases
1. 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
2. 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.
3. The letters of the word RIGHTEOUSLY are arranged in a row at random. Find the probability that the even places are filled with consonants.
4. 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
5. The letters of the word TROPICAL are arranged in a row at random. Find the probability that no two vowels may come together.
Ans : 1; 43,200; 18,14,400; 7,200; 14,400

# Solution 1

Number of Letters/Characters

= 5

{A, B, E, L, T}

⇒ nL = 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.

= 1P1 × 1P1 × ...

= 1 × 1 × ...

= 1

⇒ mA = 1

# Solution 2

In the word SLAUGHTER,

Number of Letters/Characters

= 9

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

⇒ nL = 9

Number of places to be filled in forming the required words

⇒ Number of letters in the word to be formed

= 9

⇒ nP = 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, _}

⇒ nSP = 5

Number of specified letters

⇒ Number of Vowels

= 3

{A, U, E}

⇒ nSL = 3

After filling the even places with vowels

Number of places filled

= 3

Smaller of nSL (3) and nSP (5)

⇒ nFP = 3

Number of Places remaining to be filled

= Total Places − Places Filled

 ⇒ nRP = nP − nFP = 9 − 3 = 6

Number of Letters remaining to be used

= Total Letters − Places Filled

 ⇒ nRL = nL − nFP = 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

 ⇒ mB = aPb × nRLPnRP Where a is the larger of nSL (3) and nSP (5) and b the other 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 = 5P3 × 6P6 = (5 × 4 × ... 3 terms) × 6! = (5 × 4 × 3) × (6 × 5 × 4 × 3 × 2 × 1) = 60 × 720 = 43,200

# Solution 3

In the word RIGHTEOUSLY,

Number of Letters/Characters

= 11

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

⇒ nL = 9

Number of places to be filled in forming the required words

⇒ Number of letters in the word to be formed

= 11

⇒ nP = 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}

⇒ nSP = 5

Number of specified letters

⇒ Number of Consonants

= 7

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

⇒ nSL = 7

After filling the even places with vowels

Number of places filled

= 3

Smaller of nSL (7) and nSP (5)

⇒ nFP = 5

Number of Places remaining to be filled

= Total Places − Places Filled

 ⇒ nRP = nP − nFP = 11 − 5 = 6

Number of Letters remaining to be used

= Total Letters − Places Filled

 ⇒ nRL = nL − nFP = 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

 ⇒ mH = aPb × nRLPnRP Where a is the larger of nSL (7) and nSP (5) and b the other 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 = 7P5 × 6P6 = (7 × 6 × ... 5 terms) × 6! = (7 × 6 × 5 × 4 × 3) × (6 × 5 × 4 × 3 × 2 × 1) = 2,520 × 720 = 18,14,400

# Solution 4

In the word PENTAGON,

Number of Letters/Characters

= 8

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

⇒ nL = 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

1. The First and the Fourth places

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

2. The Second and the Fifth places

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

3. The Third and the Sixth places

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

4. The Fourth and the Seventh places

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

5. 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

 ⇒ mT = 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 = 5C1 × 2P2 × 6P6 = 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 TEA1

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

 ⇒ mTEA1 = 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 = 2P2 × 6P6 = 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.

⇒ mTEA1 = mTEA1 = mTEA1 = mTEA1 = mTEA1 = 1,440

Total Number of Favorable Choices for event T

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

 ⇒ mT = ΣmTEAi = mTEA1 + mTEA1 + mTEA1 + mTEA1 + mTEA1 = 1,440 + 1,440 + 1,440 + 1,440 + 1,440 = 7,200

# Solution 5

In the word TROPICAL,

Number of Letters/Characters

= 8

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

⇒ nL = 8

Number of places to be filled in forming the required words

⇒ Number of letters in the word to be formed

= 8

⇒ nP = 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}

⇒ nDL = 3

Number of other letters

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

 ⇒ nOL = nL − nDL = 8 − 3 = 5
 nDL − 1 = 3 − 1 = 2
Since nOL ≥ nDL − 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

 ⇒ mL = 5P5 × 6P3 = 5! × (6 × 5 × ... 3 terms) = (5 × 4 × 3 × 2 × 1) × (6 × 5 × 4) = 120 × 120 = 14,400