Question

If all the permutations of the letters in the word ${\text{OBJECT}}$ are arranged in alphabetical order as in a dictionary. The ${717^{th}}$word is
$\,A\,)\,T\,J\,O\,E\,C\,B \\\ \,B\,)\,T\,O\,J\,E\,C\,B\, \\\ \,C\,)\,T\,O\,J\,C\,B\,E \\\ \,D\,)\,{\text{none}}\,{\text{of}}\,{\text{these}} \\\$

Answer 1

First calculate the total letters in the given word. After that by taking permutation. Find the first letter by applying $5\,!$ to all the words and find the second letter by applying $4\,!$to all the words and so on. If the total count exceeds$717$, stop the process and decrement the total by $1$until we get $717$as the total count.

Formula used: The factorial concept is used in this problem. The factorial of $5\,!$is $5\,!\, = \,1\, \times \,2 \times 3 \times 4 \times 5\, = \,120$.As like that $4\,!$is $4\,!\, = \,1\, \times \,2\, \times \,3\, \times \,4\, = \,24$.

Complete step-by-step answer:
Given that, all the permutations of the letters in the word $OBJECT$ are arranged in alphabetical order as in the dictionary.
The word $OBJECT$ contains $6$ letters in it.
The words are arranged in alphabetical order, the first word remains the same. Only the remaining words will be changed.
That is the total letters in the given word $OBJECT$ is $6$.
The first permutation of the given word contains $5$ letters.
Let us consider the word starts with $B$:
So, the remaining $5$ letters are involved in permutation as $5\,!$.
The value of $5\,!$ is $120.$
As similar to the letter $B$, we want to find the permutations for all the letters:
Let us consider the word starts with $B$:
So, the remaining $5$letters are involved in permutation as$5\,!$.
The value of $5\,!$ is $120.$
Let us consider the word starts with $C$:
So, the remaining $5$letters are involved in permutation as$5\,!$.
The value of $5\,!$ is $120.$
The total permutation of the letters $B$ and $C$are$120\, + \,120\, = \,240$.
Let us consider the word starts with $E$:
So, the remaining $5$letters are involved in permutation as$5\,!$.
The value of $5\,!$ is $120.$
The total permutation of the letters $B$,$C$ and $E$ are$\,240\, + \,120\, = \,360$.

Let us consider the word starts with $J$:
So, the remaining $5$letters are involved in permutation as$5\,!$.
The value of $5\,!$ is $120.$
The total permutation of the letters $B$,$C$, $E$ and $J$are$360\, + \,120\, = \,480$.
Let us consider the word starts with $O$:
So, the remaining $5$letters are involved in permutation as$5\,!$.
The value of $5\,!$ is $120.$
The total permutation of the letters $B$,$C$, $E$ and $J$ are $\,480\, + \,120\, = \,600$.
Now, the ${601^{st}}$word will start with the letter $T$.
Now, find the second occurrence letter in the word:
The word starts with $T$ and we have to assume the second letter, then the number of permutations in the word is$4\,!$.
Let us consider the second word as $B$:
So, the remaining $4$ letters are involved in permutations as$4\,!$.
The value of $4\,!$ is $24$.
Calculate the total by adding permutation total of first letter and current occurrence:
The total permutation of the letter $T$ and $B$ is$600\, + \,24\, = \,624$.
Let us consider the second word as $C$:
So, the remaining $4$ letters are involved in permutations as $4\,!$.
The value of $4\,!$ is$24$.
The total permutation of the letter $T$ and $C$ is $624\, + \,24\, = \,648$.
Let us consider the second word as $E$:
So, the remaining $4$ letters are involved in permutations as $4\,!$.
The value of $4\,!$ is$24$.
The total permutation of the letter $T$ and $E$ is $\,648\, + \,24\, = \,672$.
Let us consider the second word as $J$:
So, the remaining $4$letters are involved in permutations as$4\,!$.
The value of $4\,!$ is$24$.
The total permutation of the letter $T$ and $J$is$672\, + \,24\, = \,696$.
Let us consider the second word as $O$:
So, the remaining $4$ letters are involved in permutations as $4\,!$.
The value of $4\,!$ is$24$.
The total permutation of the letter $T$ and $O$ is $696\, + \,24\, = \,720$.
At the letters $T$ and $O$, the total value exceeded to $720$.
But we actually want to find the occurrences of ${717^{th}}$word:
So, decrement three counts from the word$720$.
We take that the ${720^{th}}$ word is $TOJECB$. This is taken because except $T$ and $O$ all the remaining words are in decreasing order.
${720^{th}}\, = \,TOJECB$
Now, we want to decrement$1$:
We take that the ${719^{th}}$ word is $TOJEBC$. This is taken by decrementing $1$ so the letter $B$ will alphabetically come before $C$.
${719^{th}}\, = \,TOJEBC$
Now, we want to decrement$1$:
We take that the ${718^{th}}$ word is $TOJCEB$. This is taken by decrementing $1$ so the letter $C$ will alphabetically come before $E$ and the letter $B$ follows $E$.
${718^{th}}\, = \,TOJCEB$
Now, we want to decrement$1$:
We take that the ${717^{th}}$ word is $TOJCBE$. This is taken by decrementing $1$ so the letter $B$ will alphabetically come before $E$ and the letter $C$ remains the same place.
${717^{th}}\, = \,TOJCBE$.
Thus, the value of ${717^{th}}$ word is $TOJCBE$.
So, the option $C\,)TOJCBE$ is the correct answer.

Note: By decrementing the value of $720$ to $717$, we have to check based on the dictionary condition. In the dictionary each letter occurs in alphabetical order like TAB will occur before TBA.