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Question: If \(^{2n + 1}{P_{n - 1}}{:^{2n - 1}}{P_n} = 3:5\), then \(n\) is:...

If 2n+1Pn−1:2n−1Pn=3:5^{2n + 1}{P_{n - 1}}{:^{2n - 1}}{P_n} = 3:5, then nn is:

Explanation

Solution

We will first write the given ratio in terms of fraction. Then, simplify the expression using the formula nPr=n!(n−r)!^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}} and n!=n.(n−1).(n−2)......3.2.1n! = n.\left( {n - 1} \right).\left( {n - 2} \right)......3.2.1. Then, cross multiply and form a quadratic equation. Factorise the quadratic equation and solve for the value of nn.

Complete step-by-step answer:
We are given that 2n+1Pn−1:2n−1Pn=3:5^{2n + 1}{P_{n - 1}}{:^{2n - 1}}{P_n} = 3:5
We can rewrite the given ratio in fraction.
2n+1Pn−12n−1Pn=35\dfrac{{^{2n + 1}{P_{n - 1}}}}{{^{2n - 1}{P_n}}} = \dfrac{3}{5}
We know that nPr=n!(n−r)!^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}
Therefore, on simplifying the expression, we will get,
(2n+1)!(2n+1−(n−1))!(2n−1)!(2n−1−(n))!=35\dfrac{{\dfrac{{\left( {2n + 1} \right)!}}{{\left( {2n + 1 - \left( {n - 1} \right)} \right)!}}}}{{\dfrac{{\left( {2n - 1} \right)!}}{{\left( {2n - 1 - \left( n \right)} \right)!}}}} = \dfrac{3}{5}
Now, we will solve the brackets and we will use the formula n!=n.(n−1).(n−2)......3.2.1n! = n.\left( {n - 1} \right).\left( {n - 2} \right)......3.2.1
(2n+1)!(2n+1−n+1)!(2n−1)!(2n−1−n)!=35 ⇒(2n+1)!(n+2)!×(n−1)!(2n−1)!=35 ⇒(2n+1)(2n)(2n−1)!(n−1)!(n+2)(n+1)n.(n−1)!(2n−1)!=35 ⇒(2n+1)2(n+2)(n+1)=35  \dfrac{{\dfrac{{\left( {2n + 1} \right)!}}{{\left( {2n + 1 - n + 1} \right)!}}}}{{\dfrac{{\left( {2n - 1} \right)!}}{{\left( {2n - 1 - n} \right)!}}}} = \dfrac{3}{5} \\\ \Rightarrow \dfrac{{\left( {2n + 1} \right)!}}{{\left( {n + 2} \right)!}} \times \dfrac{{\left( {n - 1} \right)!}}{{\left( {2n - 1} \right)!}} = \dfrac{3}{5} \\\ \Rightarrow \dfrac{{\left( {2n + 1} \right)\left( {2n} \right)\left( {2n - 1} \right)!\left( {n - 1} \right)!}}{{\left( {n + 2} \right)\left( {n + 1} \right)n.\left( {n - 1} \right)!\left( {2n - 1} \right)!}} = \dfrac{3}{5} \\\ \Rightarrow \dfrac{{\left( {2n + 1} \right)2}}{{\left( {n + 2} \right)\left( {n + 1} \right)}} = \dfrac{3}{5} \\\
Solve the brackets and cross-multiply to solve the value of nn
4n+2n2+3n+2=35 ⇒20n+10=3n2+9n+6 ⇒3n2−11n−4=0  \dfrac{{4n + 2}}{{{n^2} + 3n + 2}} = \dfrac{3}{5} \\\ \Rightarrow 20n + 10 = 3{n^2} + 9n + 6 \\\ \Rightarrow 3{n^2} - 11n - 4 = 0 \\\
Factorise the above equation.
3n2−12n+n−4=0 ⇒3n(n−4)+1(n−4)=0 ⇒(3n+1)(n−4)=0  3{n^2} - 12n + n - 4 = 0 \\\ \Rightarrow 3n\left( {n - 4} \right) + 1\left( {n - 4} \right) = 0 \\\ \Rightarrow \left( {3n + 1} \right)\left( {n - 4} \right) = 0 \\\
Equate each factor to 0 to find the value of nn
3n+1=0 n=−13  3n + 1 = 0 \\\ n = - \dfrac{1}{3} \\\
And
n−4=0 n=4  n - 4 = 0 \\\ n = 4 \\\
But, nn has to be a natural number.
Therefore, n=4n = 4

Note: Many students make mistakes by using the formula of combination and not the formula of permutation. Here, in the expression, nPr^n{P_r}, PP represents permutation and is equal to n!(n−r)!\dfrac{{n!}}{{\left( {n - r} \right)!}}. And, the formula of nCr=n!r!(n−r)!^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}, where CC represents combination.