# Problem 4

The letters of the word ARRANGE are arranged at random. Find the probability that
1. the two R's come together
2. the two R's do not come together
3. the two R's and the two A's come together
4. the two R's come together and the two A's do not come together
Ans :
 2 7
,
 5 7
,
 2 21
,
 4 21

# Solution

In the word ARRANGE

Number of letters

= 7

{A, R, R, A, N, G, E}

⇒ nL = 7

Number of Letters

of a first kind

Number of A's = 2

⇒ nLa = 2

of a second kind

Number of R's = 2

⇒ nLb = 2

Number of places in the word to be formed,

⇒ Number of Letters in the word to be formed

= 7

⇒ nP = 7

Experiment :

Forming a word using the 7 letters

Total Number of Possible Choices

= Number of words that can be formed using the 7 letters of which nLa are of one kind and nLb are of another kind

⇒ n =
 nL! nLa! × nLb!
=
 7! 2! × 2!
=
 7 × 6 × 5 × 4 × 3 × 2! (2 × 1) × 2!
= 1,260

Let

A : the event of forming the word such that the two R's are together

B : the event of forming the word such that the two R's are not together

C : the event of forming the word such that the two A's and the two R's are together

D : the event of forming the word such that the two R's are together but the two A's are not together

## For Event A

Number of groups of letters

= 1

⇒ nG = 1

Number of letters in the first group

⇒ Number of R's

= 2

⇒ nGL1 = 2

Total Number of letters considering the groups of letters as a unit

= Total number of letters − Number of letters grouped + Number of groups of letters

 ⇒ nGL = (nL − $\sum _{i=1}^{{n}_{G}}{n}_{{G}_{{L}_{i}}}$) + nG = nL − (nGL1) + nG = 7 − (2) + 1 = 6

{A, (R, R), A, N, G, E}

### In the letters with grouped letters as a unit

Number of letters

of a first kind

Number of A's = 2

⇒ nGLa = 2

### In the letters of the first group

Number of letters

of a first kind

Number of R's = 2

⇒ nGL1a = 2

Number of Favorable Choices

= Number of words that can be formed using the letters of the word ARRANGE with the two R's together

= Number of ways in which the 6 letters with the grouped letters as a unit can be arranged × Number of ways in which the 2 grouped letters can be arranged between themselves

⇒ mA =
 nGL! nGLa!
×
 nGL1! nGL1a!
=
 6! 2!
×
 2! 2!
=
 6 × 5 × 4 × 3 × 2! 2!
× 1
= 360

Probability of forming words using all the letters of the word ARRANGE with the two R's together

⇒ Probability of occurrence of Event A

=
 Number of Favorable Choices for the Event Total Number of Possible Choices for the Experiment
⇒ P(A) =
 mA n
=
 360 1,260
=
 2 7

## Odds

Number of Unfavorable Choices

= Total Number of possible choices − Number of Favorable choices

 ⇒ mAc = n − mA = 1,260 − 360 = 900

## in favor

Odds in Favor of forming the words with the two R's together

⇒ Odds in Favor of Event A

= Number of Favorable Choices : Number of Unfavorable Choices

= mA : mAc

= 360 : 900

= 2 : 5

## against

Odds against forming the words with the two R's together

⇒ Odds against Event A

= Number of Unfavorable Choices : Number of Favorable Choices

= mAc : mA

= 900 : 360

= 5 : 2

## Odds (alternative)

Probability of non-occurrence of Event A

⇒ P(Ac) = 1 − P(A)
= 1 −
 2 7
=
 7 − 2 7
=
 5 7

## in favor

Odds in Favor of forming the words with no two S's together

⇒ Odds in Favor of Event A

= Probability of occurrence of the event : Probability of non-occurrence of the event
= P(A) : P(Ac)
=
 2 7
:
 5 7
= 2 : 5

## against

Odds against forming the words with the two R's together

⇒ Odds against Event A

= Probability of non-occurrence of the event : Probability of occurrence of the event
= P(Ac) : P(A)
=
 5 7
:
 2 7
= 5 : 2

## For Event B

Number of Favorable Choices

= The number of words that can be formed using the letters of the word ARRANGE with the two R's not together

= Total number of words that can be formed using the letters of the word ARRANGE − Number of words with the two R's together

 ⇒ mB = n − mA = 1,260 − 360 = 900

Probability of forming words using all the letters of the word ARRANGE with the two R's not together

⇒ Probability of occurrence of Event B

=
 Number of Favorable Choices for the Event Total Number of Possible Choices for the Experiment
⇒ P(B) =
 mB n
=
 900 1,260
=
 5 7

## For Event B [Alternative]

Probability of forming words using all the letters of the word ARRANGE with the two R's not together

⇒ Probability of occurrence of Event B

= Probability of non-occurrence of Event A

⇒ P(B) = P(Ac)
= 1 − P(A)
= 1 −
 2 7
=
 7 − 2 7
=
 5 7

## For Event C

The problem is interpreted as the two R's are to be together as a set and the two A's are to be together as another set.

Number of groups of letters

= 2

⇒ nG = 2

Number of letters in the first group

⇒ Number of R's

= 2

⇒ nGL1 = 2

Number of letters in the second group

⇒ Number of A's

= 2

⇒ nGL2 = 2

Total Number of letters considering the groups of letters as a unit

= Total number of letters − Number of letters grouped + Number of groups of letters

 ⇒ nGL = (nL − $\sum _{i=1}^{{n}_{G}}{n}_{{G}_{{L}_{i}}}$) + nG = nL − (nGL1 + nGL2) + nG = 7 − (2 + 2) + 2 = 5

{(A, A), (R, R), N, G, E}

### In the letters of the first group

Number of letters

of a first kind

Number of R's = 2

⇒ nGL1a = 2

### In the letters of the second group

Number of letters

of a first kind

Number of A's = 2

⇒ nGL2a = 2

Number of Favorable Choices

= Number of words that can be formed using the letters of the word ARRANGE with the two R's and the two A's together

= Number of ways in which the 5 letters with the grouped letters as a unit can be arranged × Number of ways in which the 2 grouped letters in the first group can be arranged between themselves × Number of ways in which the 2 grouped letters in the second group can be arranged between themselves

⇒ mC = nGLPnGL ×
 nGL1! nGL1a!
×
 nGL2! nGL2a!
= 5P5 ×
 2! 2!
×
 2! 2!
= (5 × 4 × 3 × 2 × 1) × 1 × 1
= 120

Probability of forming words using all the letters of the word ARRANGE with the two R's and the A's together

⇒ Probability of occurrence of Event C

=
 Number of Favorable Choices for the Event Total Number of Possible Choices for the Experiment
⇒ P(C) =
 mC n
=
 120 1,260
=
 2 21

## For Event C [Alternative interpretation]

The problem is interpreted as the two R's and the two A's are to be together as a set.

Number of groups of letters

= 1

⇒ nG = 1

Number of letters in the first group

⇒ Number of R's and A's

= 4

⇒ nGL1 = 4

Total Number of letters considering the groups of letters as a unit

= Total number of letters − Number of letters grouped + Number of groups of letters

 ⇒ nGL = (nL − $\sum _{i=1}^{{n}_{G}}{n}_{{G}_{{L}_{i}}}$) + nG = nL − (nGL1) + nG = 7 − (4) + 1 = 4

{(A, A, R, R), N, G, E}

### In the letters of the first group

Number of letters

of a first kind

Number of R's = 2

⇒ nGL1a = 2

of a second kind

Number of A's = 2

⇒ nGL1b = 2

Number of Favorable Choices

= Number of words that can be formed using the letters of the word ARRANGE with the two R's and the two A's together

= Number of ways in which the 4 letters with the grouped letters as a unit can be arranged × Number of ways in which the 4 grouped letters in the first group can be arranged between themselves

⇒ mC = nGLPnGL ×
 nGL1! nGL1a! × nGL1b!
= 4P4 ×
 4! 2! × 2!
= 4! ×
 4 × 3 × 2! 2! × 2 × 1
= (4 × 3 × 2 × 1) × 6
= 144

Probability of forming words using all the letters of the word ARRANGE with the two R's and the A's together

⇒ Probability of occurrence of Event C

=
 Number of Favorable Choices for the Event Total Number of Possible Choices for the Experiment
⇒ P(C) =
 mC n
=
 144 1,260
=
 4 35

## For Event D

Number of Favorable Choices

= Number of words that can be formed using the letters of the word ARRANGE such that the two R's are together and the two A's are not together

= Number of arrangements in which the two R's are together as a unit − Number of arrangements in which the two R's are together as a unit and the two A's are together as another unit

 ⇒ mD = mA − mC = 360 − 120 = 240

Probability of forming words using all the letters of the word ARRANGE with the two R's are together and the two A's are not together

⇒ Probability of occurrence of Event D

=
 Number of Favorable Choices for the Event Total Number of Possible Choices for the Experiment
⇒ P(D) =
 mD n
=
 240 1,260
=
 4 21
Author : The Edifier