Answer

Verified

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,

$\dfrac{{}^{56}{{P}_{6+r}}}{{}^{54}{{P}_{3+r}}}=\dfrac{30800}{1}$

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 ${}^{n}{{P}_{r}}={}^{n}{{C}_{r}}\times r!$.

Recently Updated Pages

Define the following A 1 volt PD B Electric power class 12 physics CBSE

In a halfwave rectifier the rms value of the ac component class 12 physics CBSE

When light travels from a rarer to a denser medium class 12 physics CBSE

The unit of electric dipole moment is A Newton B Coulomb class 12 physics CBSE

The waveforms A and B given below are given as input class 12 physics CBSE

What is the cause of Presbyopia class 12 physics CBSE

Trending doubts

Difference Between Plant Cell and Animal Cell

Mention the different categories of ministers in the class 10 social science CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Who is the executive head of the Municipal Corporation class 6 social science CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Which monarch called himself as the second Alexander class 10 social science CBSE

Select the word that is correctly spelled a Twelveth class 10 english CBSE

Write an application to the principal requesting five class 10 english CBSE