Question

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 ${50^{th}}$ word?

Answer 1

Hint: First we calculate the total number of words starting with A, G and I then we calculate ${50^{th}}$ word, according to the dictionary.
Formula Used: Permutation is used.


Complete step by step solution:
 In Again Word Repeat of letter A is 2 times G in one time and I, N is also one time.
In dictionary Letters appear alphabetically,

Words Starting with	Representation of words	Number of words
A	$A\underbrace { - - - - }_{}$ 4 Letters in which A,G, I, N	Since we arrange 4 letters number of words$ = 4{P_4}$ $ = 4! = 4 \times 3 \times 2 \times 1 \\ \,\,\,\,\,\,\,\,\,\,\, = 24 \\ $
G	$G\underbrace { - - - - }_{}$ 4 Letters in which 2A, I, N	Since Letters are repeating number of words\[\] $ = \dfrac{{n!}}{{{P_1}!\,\,{P_2}!\,\,{P_3}!}}$ No. of letters n=4Since 2A${P_1} = 4$ No. of words $ = \dfrac{{4!}}{{2!}} \\ = \dfrac{{4 \times 3 \times 2!}}{{2!}} = 12 \\ $
I	$I\underbrace { - - - - }_{}$$ letters in which 2A, G , N	$\eta = 0$ Since 2A${P_1} = 4$Number of words $ = \dfrac{{4!}}{{2!}} \\ = \dfrac{{4 \times 3 \times 2!}}{{2!}} = 12 \\ $

Thus, Total number of words starting with \[A,\,G{\text{ }}\& {\text{ }}I{\text{ }} = 24 + 12 + 12\]
  =48
Hence, ${49^{th}}$ word will start from N I.e. $N\underline A \underline A \underline G \underline I $
$4$ remaining four rearrange according to a dictionary, Thus the
${50^{th}}$ Word is \[N\underline {\,A} \,\underline {\,A} \underline {\,I} \underline {\,G} \,\]
hence, the ${50^{th}}$ word will be \[N\underline {\,A} \,\underline {\,A} \underline {\,I} \underline {\,G} \,\]


Note: Students should keep in mind that whatever comes first in the words dictionary, we start from that letter. And, also keep in mind how many words remain.