Question

Find $n$ if ${}^{n-1}{{P}_{3}}:{}^{n}{{P}_{4}}=1:9$. \[\]

Answer 1

Hint: We use the recurrence relation on factorial for steps convert the formula for $r$ permutation of $n$ that is ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ to ${}^{n}{{P}_{r}}=n\times \left( n-1 \right)\times ...\left( n-r-1 \right)$. We use it in the left hand side of the given equation for permutation and solve for $n$. \[\]

Complete step by step answer:
We know that a recurrence relation is an equation that recursively defines a sequence or multidimensional array of values, once one or more initial terms are given; each further term of the sequence or array is defined as a function of the preceding terms. \[\]
We know that the factorial of natural numbers $n$ is the product of the first $n$ natural number. We denote the factorial of $n$ as $n!$ and it is given by
\[n!=1\times 2\times 3\times ...\times n\]
We can recursively use the above formula for 1 step as,
\[\begin{align}
  & n!=n\times \left( n-1 \right)\times \left( n-2 \right)...\times 3\times 2\times 1 \\
 & \Rightarrow n!=n\left( n-1 \right)! \\
\end{align}\]
The above equation is the recursive relation for $n!$ with initial values $0!=1$.We can recursively use the above formula for $r$ steps as,
\[n!=n\left( n-1 \right)\left( n-2 \right)...\left( n-r+1 \right)\left( n-r \right)!\]
We also know from elementary combinatorics that the number of ways we can arrange $n$ distinct objects without repetition is called permutation. The number of arrangements of $n$ distinct objects in $r$ fixed slots or position is called “$r$ permutation of $n$”, denoted as ${}^{n}{{P}_{r}}$ and is given by the formula
\[{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}\]
We recursively use the recursive relation for $n!$ in the numerator with $r$ steps in the above formal to have,
\[\begin{align}
  & {}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}=\dfrac{n\times \left( n-1 \right)\times ...\left( n-r+1 \right)\left( n-r \right)!}{\left( n-r \right)!} \\
 & \Rightarrow {}^{n}{{P}_{r}}=n\times \left( n-1 \right)\times ...\left( n-r+1 \right).....\left( 1 \right) \\
\end{align}\]
We are given in the question and proportion in permutation as ,
\[\begin{align}
  & {}^{n-1}{{P}_{3}}:{}^{n}{{P}_{4}}=1:9 \\
 & \Rightarrow \dfrac{{}^{n-1}{{P}_{3}}}{{}^{n}{{P}_{4}}}=\dfrac{1}{9} \\
\end{align}\]
We use formula (1) in numerator for $n=n-1,r=3$ see that last term in with respect to formula (1) will be $n-r+1=n-1-3+1=n-3$.We use formula (1) in the denominator for $n=n,r=4$ in see that last term in with respect to formula (1) will be $n-r+1=n-4+1=n-3$. We proceed
\[\begin{align}
  & \Rightarrow \dfrac{\left( n-1 \right)\left( n-1-1 \right)\left( n-1-3+1 \right)}{n\left( n-1 \right)\left( n-2 \right)\left( n-4+1 \right)}=\dfrac{1}{9} \\
 & \Rightarrow \dfrac{\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)}{n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)}=\dfrac{1}{9} \\
\end{align}\]
We divide numerator and denominator at the left hand of the equation by $\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)$ assuming $n\ne 0,1,2,3$ and have,
\[\begin{align}
  & \Rightarrow \dfrac{1}{n}=\dfrac{1}{9} \\
 & \therefore n=9 \\
\end{align}\]

So we obtain the values of $n$ as 9 and our assumption $n\ne 0,1,2,3$ was correct.

Note: We also find recurrence relation in Fibonacci sequence, binomial coefficients etc. We note to careful of the confusion formula for permutation from combination which is the selection $r$ objects from $n$ distinct objects as ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. We see the relation between permutation and combination as ${}^{n}{{C}_{r}}\times r!={}^{n}{{P}_{r}}$. If the objects become identical, $r$ permutation of $n$ is ${{n}^{k}}$.