Question

If ${r}<{s}\le n$ then prove that ${}^{n}{{P}_{s}}$ is divisible by ${}^{n}{{P}_{r}}$ .

Answer 1

Hint:For solving this question, we will prove the desired result indirectly by proving that $\dfrac{{}^{n}{{P}_{s}}}{{}^{n}{{P}_{r}}}$ is an integer. First, we will use the formula ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ to write $\dfrac{{}^{n}{{P}_{s}}}{{}^{n}{{P}_{r}}}=\dfrac{\left( n-r \right)!}{\left( n-s \right)!}$ . After that, we solve the given inequality and prove that, $n-r>n-s$ . Then, we will apply one of the basic concept of factorial, i.e. $\dfrac{a!}{b!}$ will be an integer if $a>b$ to prove that, $\dfrac{{}^{n}{{P}_{s}}}{{}^{n}{{P}_{r}}}$ is an integer. Then, automatically the desired result will be proved.

Complete step-by-step answer:
Given:
It is given that, ${r}<{s}\le n$ and we have to prove that ${}^{n}{{P}_{s}}$ is divisible by ${}^{n}{{P}_{r}}$ .
Now, from the above data, we can interpret that $n,p$ and $s$ are three positive integers such that $s>r$ and $n\ge s$ .
Now, as we have to prove that ${}^{n}{{P}_{s}}$ is divisible by ${}^{n}{{P}_{r}}$ so, we will prove it indirectly by proving that $\dfrac{{}^{n}{{P}_{s}}}{{}^{n}{{P}_{r}}}$ is an integer.
Now, before we proceed we should know the following formula:
${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$
Now, we will simplify the term $\dfrac{{}^{n}{{P}_{s}}}{{}^{n}{{P}_{r}}}$ by using the formula above. Then,
$\begin{aligned} & \dfrac{{}^{n}{{P}_{s}}}{{}^{n}{{P}_{r}}}=\dfrac{\left( \dfrac{n!}{\left( n-s \right)!} \right)}{\left( \dfrac{n!}{\left( n-r \right)!} \right)} \\\ & \Rightarrow \dfrac{{}^{n}{{P}_{s}}}{{}^{n}{{P}_{r}}}=\dfrac{n!}{\left( n-s \right)!}\times \dfrac{\left( n-r \right)!}{n!} \\\ & \Rightarrow \dfrac{{}^{n}{{P}_{s}}}{{}^{n}{{P}_{r}}}=\dfrac{\left( n-r \right)!}{\left( n-s \right)!}................\left( 1 \right) \\\ \end{aligned}$
Now, as it is given that, $s>r$ so, we will multiply the inequality $s>r$ by $-1$ to reverse the sign of inequality. Then,
$\begin{aligned} & s>r \\\ & \Rightarrow -s<-r \\\ \end{aligned}$
Now, we will add $n$ on both sides in the above inequality. Then,
$\begin{aligned} & -s<-r \\\ & \Rightarrow {n-s}<{n-r} \\\ \end{aligned}$
Now, from the above result, we conclude that, $n-r$ will be greater than $n-s$ .
Now, as we know that if there are two positive integers $a$ and $b$ such that, $a>b$ . Then,
$\begin{aligned} & a!=1\times 2\times 3\times 4\times \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \times \left( a-3 \right)\times \left( a-2 \right)\times \left( a-1 \right)\times a \\\ & b!=1\times 2\times 3\times 4\times \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \times \left( b-3 \right)\times \left( b-2 \right)\times \left( b-1 \right)\times b \\\ \end{aligned}$
Now, from the above result, we can clearly conclude that, $\dfrac{a!}{b!}$ will be an integer if $a>b$ .
Now, as we have proved above that, ${n-s}<{n-r}$ . Then, $\dfrac{\left( n-r \right)!}{\left( n-s \right)!}=I$ where $I$ is an integer.
Now, from equation (1) we have can write that, $\dfrac{{}^{n}{{P}_{s}}}{{}^{n}{{P}_{r}}}=\dfrac{\left( n-r \right)!}{\left( n-s \right)!}$ and above we have proved that, $\dfrac{\left( n-r \right)!}{\left( n-s \right)!}=I$ where, $I$ is an integer. Then,
$\begin{aligned} & \dfrac{{}^{n}{{P}_{s}}}{{}^{n}{{P}_{r}}}=\dfrac{\left( n-r \right)!}{\left( n-s \right)!} \\\ & \Rightarrow \dfrac{{}^{n}{{P}_{s}}}{{}^{n}{{P}_{r}}}=I \\\ \end{aligned}$
Now, from the above result, we conclude that, $\dfrac{{}^{n}{{P}_{s}}}{{}^{n}{{P}_{r}}}=I$ where, $I$ is an integer. Which clearly means that, ${}^{n}{{P}_{s}}$ is divisible by ${}^{n}{{P}_{r}}$ .
Hence, proved.

Note: Here, the student should first understand what is asked in the question and then proceed in the right direction. Moreover, we should solve the inequality ${r}<{s}\le n$ carefully and try to get a useful result like ${n-r}>{n-s}$ without any mistake. And whenever we got stuck at some point try to apply the basic concepts of the number system and factorial to prove the desired result easily without any hurdle.