Question

Find the number of words without meaning that can be made with the letters of the word ‘AGAIN’. If these words are written as in a dictionary, what will be the $ {20^{th}} $ word?

Answer 1

Hint: In the given question, we are required to find out the number of arrangements of the word ‘AGAIN’ with or without meaning. The given question revolves around the concepts of permutations and combinations. We must have an idea about the multiplications rule of counting so as to solve such questions with ease. Then, we arrange all these words in alphabetical order as in a dictionary and have to find the $ {20^{th}} $ word. So, we list the letters of the given word in alphabetical order and then arrange them.

Complete step-by-step answer:
So, we are required to find the number of ways in which the letters of the word ‘AGAIN’ can be arranged.
The number of letters in the word ‘AGAIN’ is $ 5 $ .
Now, there are two A’s in the word ‘AGAIN’. So, we have to keep this in mind while arranging the letters of the word ‘AGAIN’ so as to find the number of words.
We know that the number of ways of arranging n things out of which r things are alike is $ \left( {\dfrac{{n!}}{{r!}}} \right) $ .
So, the number of total arrangements of the letters of the word ‘AGAIN’ is $ \dfrac{{5!}}{{2!}} = \dfrac{{120}}{2} = 60 $ .
Hence, the number of words which can be made with the letters of the word ‘AGAIN’ is $ 60 $ .
Now, we arrange all these $ 60 $ words in alphabetical order as in a dictionary.
The letters of the word ‘AGAIN’ arranged in alphabetical order are A, A, G, I, N.
So, we have to find the $ {20^{th}} $ word in alphabetical order.
Now, we start fixing the first two letters in the word in alphabetical order.
If we fix the two A’s in the first two places of the word, we are left with $ 3 $ places which are to be filled with G, I and N.
So, the number of ways of filling the $ 3 $ places with G, I and N is $ 3! = 6 $ .
Hence, the number of words starting with two A’s is $ 6 $ .
Similarly, if we fill up the first two places of the word with A and G, then the remaining $ 3 $ places are to be filled with A, I and N.
So, the number of words starting with A and G is $ 3! = 6 $ .
Similarly, if we fill up the first two places of the word with A and I, then the remaining $ 3 $ places are to be filled with A, G and N.
So, the number of words starting with A and I is $ 3! = 6 $ .
Now, calculating the total number of words till now, we get,
 $ \Rightarrow 6 + 6 + 6 $
 $ \Rightarrow 18 $
Now, the next word in the alphabetical order after the $ {18^{th}} $ word starts with A and N.
So, the $ {19^{th}} $ word formed with the letters of the word ‘AGAIN’ in alphabetical order is ANAGI.
The $ {20^{th}} $ word formed with the letters of the word ‘AGAIN’ in alphabetical order is ANGAI.
Hence, the $ {20^{th}} $ word formed with the letters of the word ‘AGAIN’ in alphabetical order is ANGAI.

Note: One should know about the principle rule of counting or the multiplication rule. Care should be taken while handling the calculations. Calculations should be verified once so as to be sure of the answer. One must know that the number of ways of arranging n things out of which r things are alike is $ \left( {\dfrac{{n!}}{{r!}}} \right) $ . We must know how the words are arranged in a dictionary in order to solve these types of problems.