two pairs of repeated letters coming together,one pair appearing together and the other pair not appearing together
Problem 4
 the two R's come together
 the two R's do not come together
 the two R's and the two A's come together
 the two R's come together and the two A's do not come together
2 
7 
5 
7 
2 
21 
4 
21 
Solution
Number of letters
= 7
{A, R, R, A, N, G, E}
⇒ n_{L} = 7
Number of Letters
of a first kind
Number of A's = 2
⇒ n_{La} = 2
of a second kind
Number of R's = 2
⇒ n_{Lb} = 2
Number of places in the word to be formed,⇒ Number of Letters in the word to be formed
= 7
⇒ n_{P} = 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 n_{La} are of one kind and n_{Lb} are of another kind
⇒ n  = 
 
= 
 
= 
 
=  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
⇒ n_{G} = 1
Number of letters in the first group
⇒ Number of R's
= 2
⇒ n_{GL1} = 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
⇒ n_{GL}  =  (n_{L} − $\sum _{i=1}^{{n}_{G}}{n}_{{G}_{{L}_{i}}}$) + n_{G} 
=  n_{L} − (n_{GL1}) + n_{G}  
=  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
⇒ n_{GLa} = 2
In the letters of the first group
Number of letters
of a first kind
Number of R's = 2
⇒ n_{GL1a} = 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
⇒ m_{A}  = 
 
= 
 
= 
 
=  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
= 

⇒ P(A)  = 
 
= 
 
= 

Odds
Number of Unfavorable Choices= Total Number of possible choices − Number of Favorable choices
⇒ m_{A}^{c}  =  n − m_{A} 
=  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
= m_{A} : m_{A}^{c}
= 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
= m_{Ac} : m_{A}
= 900 : 360
= 5 : 2
Odds (alternative)
Probability of nonoccurrence of Event A
⇒ P(A^{c})  =  1 − P(A)  
=  1 −
 
= 
 
= 

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 nonoccurrence of the event  
=  P(A) : P(A^{c})  
= 
 
=  2 : 5 
against
Odds against forming the words with the two R's together⇒ Odds against Event A
=  Probability of nonoccurrence of the event : Probability of occurrence of the event  
=  P(A^{c}) : P(A)  
= 
 
=  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
⇒ m_{B}  =  n − m_{A} 
=  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
= 

⇒ P(B)  = 
 
= 
 
= 

For Event B [Alternative]
⇒ Probability of occurrence of Event B
= Probability of nonoccurrence of Event A
⇒ P(B)  =  P(A^{c})  
=  1 − P(A)  
=  1 −
 
= 
 
= 

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
⇒ n_{G} = 2
Number of letters in the first group
⇒ Number of R's
= 2
⇒ n_{GL1} = 2
Number of letters in the second group
⇒ Number of A's
= 2
⇒ n_{GL2} = 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
⇒ n_{GL}  =  (n_{L} − $\sum _{i=1}^{{n}_{G}}{n}_{{G}_{{L}_{i}}}$) + n_{G} 
=  n_{L} − (n_{GL1} + n_{GL2}) + n_{G}  
=  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
⇒ n_{GL1a} = 2
In the letters of the second group
Number of letters
of a first kind
Number of A's = 2
⇒ n_{GL2a} = 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
⇒ m_{C}  =  ^{nGL}P_{nGL} ×
 
=  ^{5}P_{5} ×
 
=  (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
= 

⇒ P(C)  = 
 
= 
 
= 

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
⇒ n_{G} = 1
Number of letters in the first group
⇒ Number of R's and A's
= 4
⇒ n_{GL1} = 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
⇒ n_{GL}  =  (n_{L} − $\sum _{i=1}^{{n}_{G}}{n}_{{G}_{{L}_{i}}}$) + n_{G} 
=  n_{L} − (n_{GL1}) + n_{G}  
=  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
⇒ n_{GL1a} = 2
of a second kind
Number of A's = 2
⇒ n_{GL1b} = 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
⇒ m_{C}  =  ^{nGL}P_{nGL} ×
 
=  ^{4}P_{4} ×
 
=  4! ×
 
=  (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
= 

⇒ P(C)  = 
 
= 
 
= 

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
⇒ m_{D}  =  m_{A} − m_{C} 
=  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
= 

⇒ P(D)  = 
 
= 
 
= 
