Courses for Kids
Free study material
Offline Centres
Store Icon

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

Last updated date: 14th Jun 2024
Total views: 392.4k
Views today: 4.92k
392.4k+ views
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,
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 $\left( 51-r \right)!=\left( 50-r \right)!\left( 51-r \right)!$.
  & \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 \\
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.
  & \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 \\
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 ${}^{n}{{P}_{r}}={}^{n}{{C}_{r}}\times r!$.