Question

If all permutations of the letters of the word AGAIN are arranged in the order as in a dictionary. What is the 49th word?

Answer 1

$\underline{\LARGE{\textsf{Concept \: Behind}}}$

$\large{\blacktriangleright}\: \normalsize\sf \textsf{In a dictionary, all the words and letters in a Word}\\\sf\textsf{follows Alphabetical Order.}$

$\large{\blacktriangleright}\: \ \normalsize \sf\textsf{If there are \textbf{n} things,}\\\sf\textsf{No. of different arrangements that can be made} \\ \sf\textsf{considering \textbf{r} things(where r \leq n) is given by}\\\qquad\bigstar \qquad \boxed{\displaystyle\Large{^{n}P_{r} = \dfrac{n\: !}{(n\: -\: r)\: !}}}$

$\large{\blacktriangleright} \: \sf\normalsize{\textsf{If there exists a total of \textbf{n} things,out of which}}\\\sf\textsf{ p_{1} things are the same of one kind;} \\ \sf\textsf{ p_{2} things are the same of another kind;}\\\sf\textsf{ p_{3} things are the same of third kind};\\\sf\textsf{and so on and}$
$\sf\textsf{ p_{r} are the same of rth kind,};\\\sf\textsf{such that } \mathbf{\left( p_{1} + p_{2} + p_{3} + ....... +p_{r}\right) = n}}$

$\sf\textsf{Then, the Number of permutations}\\\sf\textsf{of these n things is :}\\\qquad\bigstar \qquad \boxed{\Large{\dfrac{n!}{\left(p_{1}!\right)\cdot\left(p_{2}!\right)\cdot\left(p_{3}!\right)......\left(p_{r}!\right)}}}$



$\underline{\underline{\Huge{\textbf{Solution:}}}}$

$\sf\textsf{Given Word = }\Large{\mathbb{AGAIN}}$

$\underline{\Large{\textsf{Step-1:}}}$
$\sf\textsf{Note the Number of Letters and}\\\sf\textsf{Repe}\sf\textsf{ated Letters in the given letters.}$

No. of Letters in the word = 5

No. of Repeated Letters= 2
$\sf\textsf{(Two A's)}$

$\underline{\Large{\textsf{Step-2:}}}$
$\sf\textsf{Arrange the Letters of the given word}\\\sf\textsf{in Alphabetical order}$
$\textsf{(Without considering Rep}\sf\textsf{eated Letters)}$

$\maltese \quad \sf\textsf{The Order will be : \: } \large{\mathbb{A,\: G ,\: I,\: N}}$

$\underline{\Large{\textsf{Step-3:}}}$
$\sf\textsf{Find No. of Words starting with A}\\\sf\textsf{that can be arranged using the letters}\\\sf\textsf{A, G, I and N}$

No. of Words Starting with A, is equal to No. of ways of arranging the remaining Letters in 4 empty boxes, after fixing A as First letter.

$\underline{\bf \mathbb{A}} \quad \underbrace{\square \quad \square \quad \square \quad \square }_{4\: \text{letters}} \\\small{\boxed{\text{Letters available = 1 A, 1 G, 1 I, 1 N}}}$

No. of ways of arranging 4 letters in the 4 empty boxes = $\sf\textsf{ ^{4}P_{4} = 4! = 24}$

$\therefore \sf\textsf{No. of words start with A = 24}$

$\underline{\Large{\textsf{Step-4:}}}$
$\sf\textsf{Find No. of Words starting with G}\\\sf\textsf{that can be arranged using the letters}\\\sf\textsf{A, A, I and N}$

No. of Words Starting with G, is equal to No. of ways of arranging the remaining Letters in 4 empty boxes, after fixing G as First letter.

$\underline{\bf \mathbb{G}} \quad \underbrace{\square \quad \square \quad \square \quad \square }_{4\: \text{letters}} \\\small{\boxed{\text{Letters available = 2 A's, 1 I, 1 N}}}$

No. of ways of arranging 4 letters in the 4 empty boxes (with 2 letters being repeated)=$\sf\textsf{ \dfrac{4!}{2!} = 4 \times 3 = 12}$

$\therefore \sf\textsf{No. of words start with G = 12}$

$\underline{\Large{\textsf{Step-5:}}}$
$\sf\textsf{Find No. of Words starting with I}\\\sf\textsf{that can be arranged using the letters}\\\sf\textsf{A, A, I and G}$

$\underline{\bf \mathbb{I}} \quad \underbrace{\square \quad \square \quad \square \quad \square }_{4\: \text{letters}} \\\small{\boxed{\text{Letters available = 2 A's, 1 G, 1 N}}}$

No. of ways of arranging 4 letters in the 4 empty boxes (with 2 letters being repeated)=$\sf\textsf{ \dfrac{4!}{2!} = 4 \times 3 = 12}$

$\therefore \sf\textsf{No. of words start with I = 12}$

$\underline{\Large{\textsf{Step-6:}}}$
$\sf\textsf{Note the Total No. of words}$

$\textsf{Here, we have}$
$\boxed{ \rm \: \sf 48 \: words\begin{cases}24&\text{Words start with \bf{A}}\\12&\text{Words start with \bf{G}}\\12&\text{Words start with\; \bf{I}}\end{cases}}$

$\underline{\Large{\textsf{Step-7:}}}$
$\sf\textsf{Note the starting letter of 49th Word}$

$\sf\textsf{As per the Alphabetical order,}\\\sf\textsf{The 49th word Starts with N}$

$\underline{\bf \mathbb{N}} \quad \underbrace{\square \quad \square \quad \square \quad \square }_{4\: \text{letters}} \\\small{\boxed{\text{Letters available = 2 A's, 1 G, 1 I}}}$

$\underline{\Large{\textsf{Step-8:}}}$
$\sf\textsf{Arrange the Remaining letters}\\\sf\textsf{in Alphabetical order}$

$\boxed{\mathbb{N}}\boxed{\mathbb{A}}\boxed{ \mathbb{A}}\boxed{ \mathbb{G}}\boxed{ \mathbb{I}}$

$\blacksquare\: \: \sf\textsf{The required 49th word is \underline{\Large{\textbf{NAAGI}}}}$

Answer 2

Step-by-step explanation:

A G A I N <=> A A G I N .

1. No. of words starting with A = 4! = 24
2. No. of words starting with G = 4!/2! = 12
3. No. of words starting with I = 4!/2! = 12

Total No of words = 48 . (i)

As we know the very next word will start with N ( Also the word includes AAGI after N )

Therefore , 49th word = NAAGI ( from i )

Thanks .........