Question

The letters of the word “INDIA” are arranged in a dictionary. Find the ${1^{st}},{13^{th}},{25^{th}},{39^{th}}$ and ${60^{th}}$ words.

Answer 1

Hint:We will first find out by the arrangement the first word and then calculate the number of words possible, when first letter is fixed to be A and similarly going on, we will see that finding the words will become easy and we will see a trick in that.

Complete step-by-step answer:
We have the word “INDIA”. Here, we have 2 I’s, 1 N, 1 D and 1 A.
We know that A comes the first in Alphabets, then D, then I and at last N.
So, the first word will be “ADIIN”.
Hence, the ${1^{st}}$ word is “ADIIN”.
Now, let us fix the first letter to be A, then we will get something like “A _ _ _ _”.
We have to fill these blanks with 4 letters out of which 3 are distinct, which can be done in $\dfrac{{4!}}{{2!}}$ ways.
(We had 4 letters therefore, the numerator has 4! But 2 letters are the same, therefore the denominator has 2!)
The number of words with A fixed as the initial letter will be hence, $\dfrac{{4!}}{{2!}} = 4 \times 3 = 12$ words.
Since, we are required to find the ${13^{th}}$ word. Hence, after A, the words will start from D.
So, the first word which starts from D will be the required ${13^{th}}$ word in the dictionary.
The first word with the first letter being D will be “DAIIN”.
Hence, the ${13^{th}}$ word is “DAIIN”.
Now, considering D to be fixed, we can have something like “D _ _ _ _”.
We have to fill these blanks with 4 letters out of which 3 are distinct, which can be done in $\dfrac{{4!}}{{2!}}$ ways.
(We had 4 letters therefore, the numerator has 4! But 2 letters are the same, therefore the denominator has 2!)
The number of words with D fixed as the initial letter will be hence, $\dfrac{{4!}}{{2!}} = 4 \times 3 = 12$ words.
Since, we are required to find the ${25^{th}}$ word. Hence, after D, the words will start from I.
So, the first word which starts from I will be the required ${25^{th}}$ word in the dictionary because 12 were covered with A and then 12 with D.
The first word with first letter being I will be “IADIN”.
Hence, the ${25^{th}}$ word is “IADIN”.
Now, considering I to be fixed, we can have something like “I _ _ _ _”.
We have to fill these blanks with 4 letters out of which 4 are distinct, which can be done in $4!$ ways.
The number of words with I fixed as the initial letter will be hence, $4! = 4 \times 3 \times 2 = 24$ words.
Since, we are required to find the ${49^{th}}$ word. Hence, after I, the words will start from N.
So, the first word which starts from N will be the required ${49^{th}}$ word in the dictionary because 12 were covered with A, then 12 with D and then 24 with I.
The first word with the first letter being N will be “NADII”.
Hence, the ${49^{th}}$ word is “NADII”.
Now, we will see how many total words were possible out of it:
This will be $\dfrac{{5!}}{{2!}}$ words.
(We had 5 letters therefore, the numerator had 5! But 2 letters are the same, therefore the denominator has 2!)
This will equate to a total of $\dfrac{{5!}}{{2!}} = 5 \times 4 \times 3 = 60$ words.
Now, we need the ${60^{th}}$ which will be the last word.
Hence, the ${60^{th}}$ word is “NIIDA”, which is just the reverse of the first word.

So, the correct answer is “Option A”.

Note:The students must remember that to approach this problem, if they actually start writing all the words, it will become really difficult as there are a total of 60 words possible.The students must note that “permutation and combination” makes our tasks so easy. We cannot go on writing all the words and combinations possible out of this word. So, we have an easy way to find its number at least.