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}
⇒ 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 | = |
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= |
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= |
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= | 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 − ) + 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 | = |
| ||||
= |
| |||||
= |
| |||||
= | 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
⇒ 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 −
| |||
= |
| |||
= |
|
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 : 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 : 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
= |
|
⇒ P(B) | = |
| ||
= |
| |||
= |
|
For Event B [Alternative]
⇒ Probability of occurrence of Event B
= Probability of non-occurrence of Event A
⇒ P(B) | = | P(Ac) | ||
= | 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
⇒ 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 − ) + 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 ×
| ||||
= | 5P5 ×
| |||||
= | (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
⇒ 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 − ) + 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 ×
| ||
= | 4P4 ×
| |||
= | 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
⇒ 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
= |
|
⇒ P(D) | = |
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= |
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= |
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