Kenneth Rosen Edition 6th Exercise 5.5 Question 35 (Page No. 380)

Question

How many strings with seven or more characters can be formed from the letters of the word $\text{EVERGREEN}$ ?

Answer 1

EVERGREEN $\Rightarrow \big[\text{E repeated 4 times }\big],\big[\text{R repeated 2 times }\big],\big[\text{V repeated 1 times }\big],\big[\text{G repeated 1 times }\big],\big[\text{N repeated 1 times }\big]$

$\color{violet}{\text{Now, the question is How many strings with seven or more characters can be formed}}$

$\color{violet}{\text{ from the letters of the word}}$ $\color{red}{'EVERGREEN'}$

$\color{blue}{\text{1. Seven letter string  :}}$

• $\color{chocolate}{\text{Forming word except the letter}}$ $\color{maroon}{\{V,G\}}\rightarrow \color{red}{ EEEERRN}$ or $\color{maroon}{\{G,N\}} \rightarrow \color{red}{EEEERRV}$ or $\color{maroon}{\{V,N\}} \rightarrow \color{red}{EEEERRG}$

          $\text{Number of ways will be}$ $\Bigg[\dfrac{7!}{4!\times 2!} + \dfrac{7!}{4!\times 2!} + \dfrac{7!}{4!\times 2!}\Bigg] = 3 \times \dfrac{7!}{4!\times 2!} = 3 \times 105 = 315 $

• $\color{chocolate}{\text{Forming word except the letter}}$ $\color{maroon}{\{1R,G\}}\rightarrow \color{red}{EEEERVN}$ or $\color{maroon}{\{1R,V\}}\rightarrow \color{red}{EEEERGN}$ or $\color{maroon}{\{1R,N\}}\rightarrow \color{red}{EEEERVN}$

          $\text{ Number of ways will be}$ $\Bigg[\dfrac{7!}{4!} + \dfrac{7!}{4} + \dfrac{7!}{4}\Bigg] = 3 \times \dfrac{7!}{4} = 3 \times 210 = 630$

• $\color{chocolate}{\text{Forming word except the letter}}$ $\color{maroon}{\{1E,V\}}\rightarrow \color{red}{EEERRGN}$ or $\color{maroon}{\{1E,G\}}\rightarrow \color{red}{EEERRVN}$ or $\color{maroon}{\{1E,N\}}\rightarrow \color{red}{EEERRVG}$

           $\text{Number of ways will be}$ $\Bigg[\dfrac{7!}{2!\times 3!} + \dfrac{7!}{2!\times 3!} + \dfrac{7!}{2!\times 3!}\Bigg] = 3 \times \dfrac{7!}{2!\times 3!} = 3 \times 420 = 1260 $

• $\color{chocolate}{\text{Forming word except the letter}}$ $\color{maroon}{\{1E,1R\}}\rightarrow \color{red}{EEERVGN}$

           $\text{Number of ways will be}$ $\Bigg[\dfrac{7!}{3!}\Bigg]= 840 $

• $\color{chocolate}{\text{Forming word except 2 R's}}\rightarrow \color{red}{EEEEVGN}$

           $\text{Number of ways will be}$ $\Bigg[\dfrac{7!}{4!}\Bigg]=  210$

• $\color{chocolate}{\text{Forming word except 2 E's}}\rightarrow \color{red}{EERRVGN}$

           $\text{Number of ways will be}$ $\Bigg[\dfrac{7!}{2! \times 2!}\Bigg]= 1260 $

$\color{green}{\text{Total Number of ways of forming 7 letter word =}}$ $(315+630+1260+840+210+1260) = \color{lightblue}{4515}$

2. $\color{blue}{\text{Eight letter string  :}}$

• $\color{chocolate}{\text{Forming word except 1 R}}\rightarrow \color{red}{EEEERVGN}$

           $\text{Number of ways will be}$ $\Bigg[\dfrac{8!}{4!}\Bigg]= 1680  $

• $\color{chocolate}{\text{Forming word except 1 E}}\rightarrow \color{red}{EEERRVGN}$

           $\text{Number of ways will be}$ $\Bigg[\dfrac{8!}{3! \times 2!}\Bigg]=  3360$

• $\color{chocolate}{\text{Forming word except}}$ $\color{maroon}{V} \rightarrow \color{red}{EEEERRGN}$ or $\color{maroon}{G} \rightarrow \color{red}{EEEERRVN}$ or $\color{maroon}{N} \rightarrow \color{red}{EEEERRVG}$

           $\text{Number of ways will be}$ $\Bigg[\dfrac{8!}{4! \times 2!} + \dfrac{8!}{4! \times 2!} + \dfrac{8!}{4! \times 2!}\Bigg] =3 \times \dfrac{8!}{4! \times 2!} = 3 \times 840 = 2520 $

$\color{green}{\text{Total Number of ways of forming 8 letter word =}}$ $(1680+3360+2520) = \color{lightblue}{7560}$

3. $\color{blue}{\text{Nine letter word :}}$

$\qquad\color{green}{\text{Number of ways will be}}$ $\Bigg[\dfrac{9!}{4! \times 2!}\Bigg]= \color{lightblue}{7560} $

∴ $\color{orange}{\text{Total number of ways = }}$ $(4515 + 7560 + 7560) =$$\color{purple}{19635}$

Answer 2

ANSWER is 19635
Total 9 Letter Strings  - 

8 Letter Strings : 
Case -1: Without one E =>

Case 2: Without one R =>

Case 3: Without any non-repeated letter 3*(V,G,N) => 

7 Letter Strings: 

Case 1: Without two E's => 

Case2:Without two R's => 

Case3: Without one E and one R =>

Case 4: Without one E and some nonrepetitive letter =>

Case 5: Without one R and some non Rep Letter => 

Case 6: Without any Non Rep Letter => 

Comments

No actually I am telling
EVERGREEN is a 9 letter word
Now, select 8 letters from 9 letters in 9C8 ways
And then arrange them
but u have not done selection in any case
why?

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In all the below cases we are left with 8 letters after excluding one each then we are left with deciding n alphabets with k repetitions and how many possibilites - $\frac {n!}{k!}$
Case 1: 
VERGREEN => 8 letters 8 positions with 3 E's and 2 R's => $\frac {8!}{3!*2!}$

Case 2:
EVEGREEN => 8 letters 8 positions with 4 E's => $\frac {8!}{4!}$

Case 3:
Without any non-repeated letter 3*(V,G,N) { Strings => EERGREEN,EVERREEN,EVERGREE}
=>$\frac {3*8!}{4!}$
 

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yes, that I know

Say for example a question, among 4 balls how nany ways we can choose 3 balls?

In this question we first have to do ⁴C₃

then we can do other part of question

But here u r not selecting rather doing arrangement at first

why?