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If four times the number of permutations of $n$ things taking $3$ at a time is equal to five times the number of permutations of $n - 1$ things taking $3$ at a time, find the value of $n$.

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Hint: Use formula of permutation, $^n{P_r} = \frac{{n!}}{{(n - r)!}}.$
The number of permutations of $n$ things taking $3$ at a time \[ = {{\text{ }}^n}{P_3}.\]
Similarly, the number of permutations of $n - 1$ things taking $3$ at a time \[ = {{\text{ }}^{n - 1}}{P_3}.\]
Now, according to question:
\[ \Rightarrow 4{ \times ^n}{P_3} = 5{ \times ^{n - 1}}{P_3}\]
We know that, $^n{P_r} = \frac{{n!}}{{(n - r)!}}$ using this we’ll get:
\[
   \Rightarrow 4 \times \frac{{n!}}{{(n - 3)!}} = 5 \times \frac{{(n - 1)!}}{{(n - 4)!}} \\
   \Rightarrow \frac{{4n(n - 1)!}}{{(n - 3)(n - 4)!}} = \frac{{5(n - 1)!}}{{(n - 4)!}} \\
   \Rightarrow \frac{{4n}}{{n - 3}} = 5 \\
   \Rightarrow 4n = 5n - 15 \\
   \Rightarrow n = 15 \\
\]
Therefore, the required value of $n$ is $15$
Note: Permutation of 3 things includes both the selection of 3 things and their arrangement as well.
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