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If all the seven letters of the word LEADING are permuted in all possible ways and the words thus formed are arranged as in the dictionary order, then the word in $2017^{\text {th }}$ place is
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Verified Answer
The correct answer is:
ELNADGI
When letters of LEADING are arranged in ascending order then letters are A, D, E, G, I, L, N Now, words starts with $\mathrm{A}$ is $6 !=720$, ie. $A$........
Similarly, words starts with $D=6 !=720$
Now, words starts with $E$ is also $6 !=720$.
But total number is 2160 that is more than 2017th place. Hence, word will start with $E$.
Now, words starts with
$$
\begin{aligned}
& E A \ldots \ldots .=5 !=120 \\
& E D \ldots \ldots . .=5 !=120 \\
& E G \ldots \ldots . .=5 !=120 \\
& E I \ldots \ldots . .=5 !=120
\end{aligned}
$$
Here, total words till now is $720+480=1920$
Now, words start with
$$
\begin{aligned}
& \text { ELA } \ldots \ldots \ldots=4 !=24 \\
& E L D \ldots \ldots .=4 !=24 \\
& E L G \ldots \ldots .=4 !=24 \\
& E L I \ldots \ldots . .=4 !=24
\end{aligned}
$$
Now, total words is 2016.
Next word is 2017th word and the word is ELNADGI.
Similarly, words starts with $D=6 !=720$
Now, words starts with $E$ is also $6 !=720$.
But total number is 2160 that is more than 2017th place. Hence, word will start with $E$.
Now, words starts with
$$
\begin{aligned}
& E A \ldots \ldots .=5 !=120 \\
& E D \ldots \ldots . .=5 !=120 \\
& E G \ldots \ldots . .=5 !=120 \\
& E I \ldots \ldots . .=5 !=120
\end{aligned}
$$
Here, total words till now is $720+480=1920$
Now, words start with
$$
\begin{aligned}
& \text { ELA } \ldots \ldots \ldots=4 !=24 \\
& E L D \ldots \ldots .=4 !=24 \\
& E L G \ldots \ldots .=4 !=24 \\
& E L I \ldots \ldots . .=4 !=24
\end{aligned}
$$
Now, total words is 2016.
Next word is 2017th word and the word is ELNADGI.
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