Question: Find the number of words which can be made using all the letters of the word AGAIN in a dictionary o...

Question

Find the number of words which can be made using all the letters of the word AGAIN in a dictionary order. What will be the fiftieth word?

Answer 1

Solution

We first use the permutation rule to find the number of words which can be made using all the letters of the word AGAIN. Then we find the dictionary order of the word AGAIN. We find the number of words which can be made using the letters using the order. We get as close t0 50 using permutation. Then we proceed using one by one-word change.

Complete step by step answer:
We need to find the number of words that can be made using all the letters of the word AGAIN. There are 5 letters in the word, 2 out of them being similar.
The number of combinations for n things out of which r are one of a kind is $\dfrac{n!}{r!}$ .
So, in case of AGAIN we have $\dfrac{5!}{2!}=\dfrac{120}{2}=60$ .
Now we have to find the fiftieth word of the arrangement if all those 60 words are arranged in the dictionary order.
So, in AGAIN the dictionary order of the letters are $A-G-I-N$ .
So, we find the number of terms of the first set of words with A being the starting letter.
The starting letter is fixed. We have left 4 distinct letters. They can be arranged in $4!=24$ ways. We still haven’t reached 50. So, we need to continue.
Then comes G as the starting number. So, G is fixed. We have left 4 letters, 2 being one of a kind. They can be arranged in $\dfrac{4!}{2!}=12$ ways.
The total number becomes $24+12=36$ . We still haven’t reached 50. So, we need to continue.
Then comes “I” as the starting number. So, “I” is fixed. We have left 4 letters, 2 being one of a kind. They can be arranged in $\dfrac{4!}{2!}=12$ ways.
The total number becomes $24+12+12=48$ . We still haven’t reached 50. So, we need to continue.
We need 2 more to reach 50. So, we start the word with N. N gets fixed.
Now according to the dictionary order $A-G-I-N$ , we go with $A-G-I$ to get NAAGI as the ${{49}^{th}}$ word. This time AA got fixed.
Then comes NAAIG as the ${{50}^{th}}$ word.

Therefore, the ${{50}^{th}}$ word is NAAIG.

Note: We need to remember that the permeation of those two similar letters ‘A’ will not be considered. The dictionary arrangement is done keeping the letters intact from the starting in order. The permutation happens from the rear side letters.