Solveeit Logo
Login

Question

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

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

Explanation

Solution

We will first start by using the property of nPr{}^{n}{{P}_{r}} that is nPr=n!(n−r)!{}^{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 308001\dfrac{30800}{1} to find the value of r.

Complete step-by-step answer:
Now, we have been given that,
56P6+r54P3+r=308001\dfrac{{}^{56}{{P}_{6+r}}}{{}^{54}{{P}_{3+r}}}=\dfrac{30800}{1}
Now, we know that the value of nPr=n!(n−r)!{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}. So, using this we will expand the terms of 56P6+r54P3+r=308001\dfrac{{}^{56}{{P}_{6+r}}}{{}^{54}{{P}_{3+r}}}=\dfrac{30800}{1} as below,
56!(56−6−r)!54!(54−3−r)!=308001\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.
56!(50−r)!54!(51−r)!=308001\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.
56!×(51−r)!(50−r)!×54!=308001\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!=(n−1)!×n!n!=\left( n-1 \right)!\times n! and cancelling the same terms in numerator and denominator.
⇒55×56×(51−r)!(50−r)!=308001\Rightarrow \dfrac{55\times 56\times \left( 51-r \right)!}{\left( 50-r \right)!}=\dfrac{30800}{1}
Now, we know that n!=(n−1)!×nn!=\left( n-1 \right)!\times n. So, we can write (51−r)!=(50−r)!(51−r)!\left( 51-r \right)!=\left( 50-r \right)!\left( 51-r \right)!.
⇒55×56×(50−r)!(51−r)(50−r)!=308001 55×56×(51−r)=30800 \begin{aligned} & \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{aligned}
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.
(51−r)=3080055×56 ⇒(51−r)=56056 ⇒51−r=10 ⇒51−10=r ⇒r=41 \begin{aligned} & \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{aligned}
So, the value of r is 41.

Note: It is important to note that we have used the fact that n!=(n−1)nn!=\left( n-1 \right)n to solve the ratio (51−r)!(50−r)!\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 nPr=nCr×r!{}^{n}{{P}_{r}}={}^{n}{{C}_{r}}\times r!.