Question

If ${}^{56}{{P}_{r+6}}:{}^{54}{{P}_{r+3}}=\left( 30800:1 \right)$, find r.

Answer 1

Hint: We will first start by using the property of ${}^{n}{{P}_{r}}$ that is ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$. Then, we will use this property to expand the terms and further simplify the expression. Then, finally we will equate it to $\dfrac{30800}{1}$ to find the value of r.

Complete step-by-step answer:
Now, we have been given that,
$\dfrac{{}^{56}{{P}_{6+r}}}{{}^{54}{{P}_{3+r}}}=\dfrac{30800}{1}$
Now, we know that the value of ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$. So, using this we will expand the terms of $\dfrac{{}^{56}{{P}_{6+r}}}{{}^{54}{{P}_{3+r}}}=\dfrac{30800}{1}$ as below,
$\dfrac{\dfrac{56!}{\left( 56-6-r \right)!}}{\dfrac{54!}{\left( 54-3-r \right)!}}=\dfrac{30800}{1}$
Now, we will solve the denominator of the both the expression in numerator and denominator.
$\dfrac{\dfrac{56!}{\left( 50-r \right)!}}{\dfrac{54!}{\left( 51-r \right)!}}=\dfrac{30800}{1}$
Now, we will simplify the left hand side of the equation.
$\dfrac{56!\times \left( 51-r \right)!}{\left( 50-r \right)!\times 54!}=\dfrac{30800}{1}$
Now, we will solve the numerator and denominator by expanding the numerator and denominator using $n!=\left( n-1 \right)!\times n!$ and cancelling the same terms in numerator and denominator.
$\Rightarrow \dfrac{55\times 56\times \left( 51-r \right)!}{\left( 50-r \right)!}=\dfrac{30800}{1}$
Now, we know that $n!=\left( n-1 \right)!\times n$. So, we can write $\left( 51-r \right)!=\left( 50-r \right)!\left( 51-r \right)!$.
$\begin{align}
  & \Rightarrow \dfrac{55\times 56\times \left( 50-r \right)!\left( 51-r \right)}{\left( 50-r \right)!}=\dfrac{30800}{1} \\
 & 55\times 56\times \left( 51-r \right)=30800 \\
\end{align}$
Now, we will simplify the equation further by taking the constant multiplication terms in left side to division in right side and solve it further to find the value of r.
$\begin{align}
  & \left( 51-r \right)=\dfrac{30800}{55\times 56} \\
 &\Rightarrow \left( 51-r \right)=\dfrac{560}{56} \\
 &\Rightarrow 51-r=10 \\
 &\Rightarrow 51-10=r \\
 &\Rightarrow r=41 \\
\end{align}$
So, the value of r is 41.

Note: It is important to note that we have used the fact that $n!=\left( n-1 \right)n$ to solve the ratio $\dfrac{\left( 51-r \right)!}{\left( 50-r \right)!}$ . The students must make sure to use this fact accurately, only then they will be able to cancel off terms and simplify further. Also, it is advisable to remember that ${}^{n}{{P}_{r}}={}^{n}{{C}_{r}}\times r!$.