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 not different

All 4

Problem 5

Define the Event and identify the number of favourable choices in the following cases

What is probability that 4 I’s appear consecutively in the word MISSISSIPPI assuming that the letters are arranged at random
Find the probability that in a random arrangement of the letters of the word UNIVERSITY, the two I’s do not come together.
The letters of the word ACCOUNTANCY are arranged in a row at random. Find the probability that (i) all c’s come together and (ii) all c's come together and all a's may not come together.
When all the letters of the word EAGLET are arranged in a row, that probability of getting a word in which E’s occupy central position is

Solution Ans :

Solution 1

In the word MISSISSIPPI

Number of letters

= 11

{M, I, S, S, I, S, S, I, P, P, I}

⇒ n_L = 11

Number of Letters

of a first kind

Number of I's = 4

⇒ n_{L_a} = 4

of a second kind

Number of S's = 4

⇒ n_{L_b} = 4

of a third kind

Number of P's = 2

⇒ n_{L_c} = 2

Number of places in the word to be formed,

⇒ Number of Letters in the word to be formed

= 11

⇒ n_P = 11

Experiment :

Forming a word using the 11 letters

Total Number of Possible Choices

= Number of words that can be formed using the 11 letters of which n_{L_a} are of one kind, n_{L_b} are of another kind and n_{L_c} are of another kind

⇒ n

n_L!

n_{L_a}! × n_{L_b}! × n_{L_c}!

11!

4! × 4! × 2!

Let

A : the event of forming the word such that the all I's are together

For Event A

Number of groups of letters

= 1

⇒ n_G = 1

Number of letters in the first group

⇒ Number of I'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
	=	11 − (4) + 1
	=	8

{M, (I, I, I, I), S, S, S, S, P, P}

In the letters with grouped letters as a unit

Number of letters

of a first kind

Number of S's = 4

⇒ n_{GL_a} = 4

of a second kind

Number of P's = 2

⇒ n_{GL_b} = 2

In the letters of the first group

Number of letters

of a first kind

Number of I's = 4

⇒ n_{GL_{1_a}} = 4

Number of Favorable Choices

= Number of words that can be formed using the letters of the word MISSISSIPPI with all I's together

= Number of ways in which the 8 letters with the grouped letters as a unit can be arranged × Number of ways in which the 4 grouped letters can be arranged between themselves

⇒ m_A

n_GL!

n_{GL_a}! × n_{GL_b}!

n_GL₁!

n_{GL_{1_a}}!

4! × 2!

8 × 7 × 6 × 5 × 4!

4! × 2 × 1

× 1

840

Solution 2

In the word UNIVERSITY

Number of letters

= 10

{U, N, I, V, E, R, S, I, T, Y}

⇒ n_L = 10

Number of Letters

of a first kind

Number of I's = 2

⇒ n_{L_a} = 2

Number of places in the word to be formed,

⇒ Number of Letters in the word to be formed

= 10

⇒ n_P = 10

Experiment :

Forming a word using the 10 letters

Total Number of Possible Choices

= Number of words that can be formed using the 10 letters of which n_{L_a} are of a kind

⇒ n

n_L!

n_{L_a}!

10!

10 × 9!

2 × 1

5 × 9!

Let

K : the event of forming the word such that the two I's are not together

The event where the two I's are not together is a complimentary of the event of the two I's being together.

We find the favorable choices for event K through the favorable choices for event K^c.

For Event K^c

Number of groups of letters

= 1

⇒ n_G = 1

Number of letters in the first group

⇒ Number of I'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
	=	10 − (2) + 1
	=	9

{U, N, (I, I), V, E, R, S, T, Y}

In the letters of the first group

Number of letters

of a first kind

Number of I's = 2

⇒ n_{GL_{1_a}} = 2

Number of Favorable Choices

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

= Number of ways in which the 9 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_K^c

^n_GLP_{n_GL} ×

n_GL₁!

n_{GL_{1_a}}!

⁹P₉ ×

9! × 1

For Event K

Number of Favorable Choices

= Number of words that can be formed using the letters of the word UNIVERSITY such that the two I's are not together

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

⇒ m_K	=	n − m_K^c
	=	(5 × 9!) − 9!
	=	(5 − 1) × 9!
	=	4 × 9!

For Event K (alternative)

Number of Letters to stay separate

⇒ Number of I's

= 2

⇒ n_DL = 2

Number of other letters

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

⇒ n_OL	=	n_L − n_DL
	=	10 − 2
	=	8

Number of places to place the letters to stay separate

= Number of other letters + 1

⇒ n_DP	=	n_OL + 1
	=	8 + 1
	=	9

Since n_DP ≥ n_DL, it would be possible to arrange the letters in such a way that the vowels do not come together.

In letters to be separated

In arranging I's in the specified places, n_DP > n_DL and there are repeating letters. Consider space to be a letter and assume that there is word formed with the letters and spaces.

Number of Letters

of a first kind

Number of I's = 2

⇒ n_{DL_a} = 2

of a second kind

Number of spaces = 7

⇒ n_{DL_b} = 7

Number of Favorable Choices

= Number of words that can be formed using the letters of the word UNIVERSITY with no two I's together

= Number of ways in which the n_OL Other letters can be arranged in as many places creating the n_DP spaces on either side for placing the n_DL letters to be kept separated × Number of ways in which the n_DL letters to be kept separated can be arranged in the n_DP spaces created

⇒ m_K

^n_OLP_{n_OL} ×

n_DP!

n_{DL_a}! × n_{DL_b}!

⁸P8 ×

2! × 7!

8! ×

9 × 8 × 7!

(2 × 1) × 7!

8! × 9 × 4

4 × 9 × 8!

4 × 9!

Solution 3

Number of Letters/Characters in the word "ACCOUNTANCY"

	=	11 {A, C, C, O, U, N, T, A, N, C, Y}
⇒ n_L	=	11

Number of Letters :

a First Kind ⇒ Number ofA's = 2 ⇒ a = 2
a Second Kind ⇒ Number ofC's = 3 ⇒ b = 3
a Third Kind ⇒ Number ofN's = 2 ⇒ c = 2
which are all different = 4 {O, U, T, Y} ⇒ x = 4
n_L = a + b + c + x

Total Number of Possible Choices

Number of words that can be formed using the 11 letters of the word
"ACCOUNTANCY"

⇒ n

n_L

a! × b! × c!

11!

2! × 3! × 2!

11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3!

2 × 1 × 3! × 2 × 1

11 × 10 × 9 × 8 × 7 × 6 × 5

16,63,200

Let "Y" and "Q" be the events of forming the words such that all "C's" come together and all "C's" come together and all "A's" may not come together

For Event "y"

Taking all "C's" as a unit

Number of Letters :

in total ⇒ n_L(s) = 9 {A, (C, C, C), O, U, N, T, A, N, Y}
of the first kind ⇒ Number ofA's = 2 ⇒ a = 2
of the second kind ⇒Number ofN's = 2 ⇒ c = 2
which are all different = 5 {(C, C, C), O, U, T, Y} ⇒ x = 4

Number of Favorable Choices

The number of words that can be formed using the letters of the word
"ACCOUNTANCY" such that all "C's" are together

⇒ m_Y

(Number of ways in which the 9 letters taking all "C's" as a unit can be arranged) × (Number of ways in which all the "C's" can be inter arranged between themselves)

2! × 2!

9 × 8 × 7 × 6 × 5 × 4 × 3 × 2!

2 × 1 × 2!

× 1

9 × 8 × 7 × 6 × 5 × 2 × 3

90,720

For Event "Q"

Taking all "C's" as a unit and all "A's" as a unit

Number of Letters :

in total ⇒ n_L(s) = 8 {(A, A), (C, C, C), O, U, N, T, N, Y}
of the first kind ⇒ Number ofN's = 2 ⇒ a = 2
which are all different = 6 {(A, A), (C, C, C), O, U, T, Y} ⇒ x = 4

Number of Favorable Choices

The number of words that can be formed using the letters of the word
"ACCOUNTANCY" such that all "C's" are together and all the "A's" are not together

⇒ m_Q

(Number ofwords that can be formed with the letters of the word "ACCOUNTANCY" with all "C's" together) − (Number ofwords that can be formed with the letters of the word "ACCOUNTANCY" with all "C's" together and all "A's" together)

(m_Y) − { (Number of ways in which the 6 letters with all "C's" as a unit and all "A's" as a unit can be arranged) × (Number of ways in which the "C's" can be inter arranged between themselves) × (Number of ways in which the "A's" can be inter arranged between themselves) }

90,720 − {

}

90,720 − {

6 × 5 × 4 × 3 × 2!

× 1 × 1

}

90,720 − 360

90,360

Solution 4

In the word EAGLET

Number of letters

= 6

{E, A, G, L, E, T}

⇒ n_L = 6

Number of Letters

of a first kind

Number of E's = 2

⇒ n_{L_a} = 2

Number of places in the word to be formed,

⇒ Number of Letters in the word to be formed

= 6

⇒ n_P = 6

Experiment :

Forming a word using the 10 letters

Total Number of Possible Choices

= Number of words that can be formed using the 10 letters of which n_{L_a} are of a kind

⇒ n

n_L!

n_{L_a}!

6 × 5!

2 × 1

3 × 5!

Let

R : the event of forming the word such that the E's occupy the middle positions

For Event R

Number of specified letters

⇒ E's

= 2

⇒ n_SL = 2

Number of specified places

⇒ Number of Middle places

= 2

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

⇒ n_SP = 4

After filling the middle places with E's

Number of places filled

= 2

Smaller of n_SL (2) and n_SP (2)

⇒ n_FP = 2

Number of Places remaining to be filled

= Total Places − Places Filled

⇒ n_RP	=	n_P − n_FP
	=	6 − 2
	=	4

Number of Letters remaining to be used

= Total Letters − Places Filled

⇒ n_RL	=	n_L − n_FP
	=	6 − 2
	=	4

In the specified letters

Number of letters

of a first kind

Number of E's = 2

⇒ n_{SL_a} = 2

Number of Favorable Choices

= Number of words that can be formed with the letters of the word EAGLET such the E's occupy the middle positions

= Number of ways in which the 2 E's can be fixed in the middle two positions × Number of ways in which the remaining 4 places can be filled with the remaining 4 letters

⇒ m_R

n_SL!

n_{SL_a}!

× ^n_SLP_{n_SL}

× ⁴P₄

1 × 4!

4 × 3 × 2 × 1

Author : The Edifier

All

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 not different

Problem 5

Solution 1

For Event A

In the letters with grouped letters as a unit

In the letters of the first group

Solution 2

For Event Kc

In the letters of the first group

For Event K

For Event K (alternative)

In letters to be separated

Solution 3

For Event "y"

For Event "Q"

Solution 4

For Event R

In the specified letters

For Event K^c