Question

If all permutations of the letters of the word ‘AGAIN’ are arranged as in the dictionary, then find the fiftieth word.
A. NAAGI
B. NAGAI
C. NAAIG
D. NAIAG

Answer 1

Hint: First we will find the number of words that start with A by using the alphabet of the word AGAIN. Similarly, we will find the number of words that start with G and I by using the alphabet of the word AGAIN. Then add all number of words and compute the fiftieth word.

Formula used:
The number of ways to arrange of \[n\] alphabet is \[n!\].
The number of ways to arrange of \[n\] alphabets where two alphabets repeated \[a\] times and \[b\] times is \[\dfrac{{n!}}{{a!b!}}\].
\[n! = n \times \left( {n - 1} \right) \times \cdots \times 3 \times 2 \times 1\]

Complete step by step solution:
In the word AGAIN, the number of alphabets is 5, and A repeated twice.
We are making a group of two A.
Now the number of alphabets is 4.
The number of words that are start with A is \[4!\].
Now applying the factorial formula:
\[4! = 4 \times 3 \times 2 \times 1\]
\[ \Rightarrow 4! = 24\]
The number of words that are start with I is \[\dfrac{{4!}}{{2!}}\].
Now applying the factorial formula:
\[\dfrac{{4!}}{{2!}} = \dfrac{{4 \times 3 \times 2 \times 1}}{{2 \times 1}}\]
Cancel out \[2 \times 1\] from denominator and numerator.
\[ \Rightarrow \dfrac{{4!}}{{2!}} = 4 \times 3\]
\[ \Rightarrow \dfrac{{4!}}{{2!}} = 12\]
The number of words that are start with G is \[\dfrac{{4!}}{{2!}}\].
Now applying the factorial formula:
\[\dfrac{{4!}}{{2!}} = \dfrac{{4 \times 3 \times 2 \times 1}}{{2 \times 1}}\]
Cancel out \[2 \times 1\] from denominator and numerator.
\[ \Rightarrow \dfrac{{4!}}{{2!}} = 4 \times 3\]
\[ \Rightarrow \dfrac{{4!}}{{2!}} = 12\]
Total number of words that are start with A, I, and G is \[24 + 12 + 12 = 48\].
The first word that start with N is NAAGI.
So, the forty ninth word is NAAGI.
Then fiftieth word is NAAIG.
Hence option C is the correct option.

Note: The number of ways to arrange of \[n\] objects where two object repeated \[a\] times and \[b\] times is \[\dfrac{{n!}}{{a!b!}}\]. If \[n\] objects out of which 2 are the same and the 2 same objects do not come together out, then the number of ways to arrange \[n\] is \[\dfrac{{\left( {n - 1} \right)!}}{{2!}}\]. The second formula is used to calculate the number of words that start with I and G.