Question

If the letters of the word BRING are permuted in all the possible ways and the words thus formed are arranged as in dictionary order, then find the ${{59}^{th}}$ word.

Answer 1

Hint: First arrange the letters of the word BRING in alphabetical order. Find the number of words that can be formed by fixing the letter B. at first place now, in the next step find the number of words that can be formed by fixing the letter G at first place. Now fix the first two letters as IB and then find the number of possible words that can be formed. Now, by fixing IG as the first two letters find the ${{60}^{th}}$ word that will be formed and hence find the ${{59}^{th}}$ word by observing the arrangement of letters in the ${{60}^{th}}$ word.
Here we have been provided with the word BRING which is to be permuted in all the possible ways and the words are arranged in the order as present in the dictionary. We have to find the ${{59}^{th}}$ words.

Complete step by step answer:
Now first let us arrange the letters of this word in the alphabetical order. So, we have BRING as the word arranged with letters in the alphabetical; order. Therefore, BRING is the first word according to dictionary order.
Let us fix the B at first Place. So, we have four letters left which is to be arranged at four places so, we have,
$\begin{matrix}
   \begin{matrix}
   \begin{matrix}
   B & \_ \\
\end{matrix} & \_ \\
\end{matrix} & \_ & \_ \\
\end{matrix}$
Number of words that can be formed with first letters as $B=4!=24$
Now let us fix the letter G at first place. So, we have again four letters left which are to be arranged at four places. So, we have,
$\begin{matrix}
   \begin{matrix}
   \begin{matrix}
   G & \_ \\
\end{matrix} & \_ \\
\end{matrix} & \_ & \_ \\
\end{matrix}$
Number of words that can be formed with first letters as $G=4!=24$
Proceeding further, let us fix the first two letters of the word with IB. so, we have three letters left that are to be arranged at three places. Therefore, we get,
$\begin{matrix}
   \begin{matrix}
   \begin{matrix}
   I & B \\
\end{matrix} & \_ \\
\end{matrix} & \_ & \_ \\
\end{matrix}$
Number of words that can be formed with first letters as $IB=3!=6$
So, the total number of words formed till now $24+24+6=54$
We have to reach $59$, so let us move a step ahead.
Now let us fix the two letters of the word with IG. So, again we have three letters left that are to be arranged at three places. Therefore, we get,
$\begin{matrix}
   \begin{matrix}
   \begin{matrix}
   I & G \\
\end{matrix} & \_ \\
\end{matrix} & \_ & \_ \\
\end{matrix}$
Number of words that can be formed with first letters as $IB=3!=6$
Therefore, the total number of words now formed = $54+6=60$. So, according to the dictionary this ${{60}^{th}}$ word will be IGRNB.
Here in this word, we can see that the letter N appears before the letter B, so, in the ${{59}^{th}}$ word the letter B must have appeared before the letter N according to the dictionary.

Therefore, the ${{59}^{th}}$ word will be IGRBN, which is our answer.

Note: One may note that we have not considered all the words that can be formed with first letter as I because if we will do so then total number of words will become $72$ which will be of no use as we have to come back to ${{59}^{th}}$ word. You must proceed like we did in the above solution. Remember that sometimes we will be given with a formed word and we will be asked the rank of that word as arranged in the dictionary. In that case also we have to proceed with the similar process used in the above solution.