Find the number of words with or without mean | vedclass

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Question

(A) The word $AGAIN$ contains $5$ letters,where $A$ appears $2$ times. The total number of words is $\frac{5!}{2!} = 60$.To find the $50^{	ext{th}}$

Vedclass · Accepted Answer

$NAAIG$

Vedclass · Answer

(A) The word $AGAIN$ contains $5$ letters,where $A$ appears $2$ times. The total number of words is $\frac{5!}{2!} = 60$.To find the $50^{	ext{th}}$ word,we arrange the letters in alphabetical order: $A, A, G, I, N$.$1$. Words starting with $A$: Fix $A$ at the first position. The remaining letters are $A, G, I, N$. Number of arrangements $= 4! = 24$.$2$. Words starting with $G$: Fix $G$ at the first position. The remaining letters are $A, A, I, N$. Number of arrangements $= \frac{4!}{2!} = 12$.$3$. Words starting with $I$: Fix $I$ at the first position. The remaining letters are $A, A, G, N$. Number of arrangements $= \frac{4!}{2!} = 12$.Total words so far $= 24 + 12 + 12 = 48$.$4$. The $49^{	ext{th}}$ word starts with $N$. Remaining letters are $A, A, G, I$. Arranging them in alphabetical order: $AA GI, AA IG, AG AI, AG IA, AI AG, AI GA, ...$$49^{	ext{th}}$ word: $NAAGI$$50^{	ext{th}}$ word: $NAAIG$

Find the number of words with or without meaning which can be made using all the letters of the word $AGAIN$. If these words are written as in a dictionary,what will be the $50^{\text{th}}$ word?

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