
If $^{2n + 1}{P_{n - 1}}{:^{2n - 1}}{P_n} = 3:5$, then $n$ is:
Answer
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Hint: We will first write the given ratio in terms of fraction. Then, simplify the expression using the formula $^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$ and $n! = n.\left( {n - 1} \right).\left( {n - 2} \right)......3.2.1$. Then, cross multiply and form a quadratic equation. Factorise the quadratic equation and solve for the value of $n$.
Complete step-by-step answer:
We are given that $^{2n + 1}{P_{n - 1}}{:^{2n - 1}}{P_n} = 3:5$
We can rewrite the given ratio in fraction.
$\dfrac{{^{2n + 1}{P_{n - 1}}}}{{^{2n - 1}{P_n}}} = \dfrac{3}{5}$
We know that $^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$
Therefore, on simplifying the expression, we will get,
$\dfrac{{\dfrac{{\left( {2n + 1} \right)!}}{{\left( {2n + 1 - \left( {n - 1} \right)} \right)!}}}}{{\dfrac{{\left( {2n - 1} \right)!}}{{\left( {2n - 1 - \left( n \right)} \right)!}}}} = \dfrac{3}{5}$
Now, we will solve the brackets and we will use the formula $n! = n.\left( {n - 1} \right).\left( {n - 2} \right)......3.2.1$
$
\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 $n$
$
\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.
$
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 $n$
$
3n + 1 = 0 \\
n = - \dfrac{1}{3} \\
$
And
$
n - 4 = 0 \\
n = 4 \\
$
But, $n$ has to be a natural number.
Therefore, $n = 4$
Note: Many students make mistakes by using the formula of combination and not the formula of permutation. Here, in the expression, $^n{P_r}$, $P$ represents permutation and is equal to $\dfrac{{n!}}{{\left( {n - r} \right)!}}$. And, the formula of $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, where $C$ represents combination.
Complete step-by-step answer:
We are given that $^{2n + 1}{P_{n - 1}}{:^{2n - 1}}{P_n} = 3:5$
We can rewrite the given ratio in fraction.
$\dfrac{{^{2n + 1}{P_{n - 1}}}}{{^{2n - 1}{P_n}}} = \dfrac{3}{5}$
We know that $^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$
Therefore, on simplifying the expression, we will get,
$\dfrac{{\dfrac{{\left( {2n + 1} \right)!}}{{\left( {2n + 1 - \left( {n - 1} \right)} \right)!}}}}{{\dfrac{{\left( {2n - 1} \right)!}}{{\left( {2n - 1 - \left( n \right)} \right)!}}}} = \dfrac{3}{5}$
Now, we will solve the brackets and we will use the formula $n! = n.\left( {n - 1} \right).\left( {n - 2} \right)......3.2.1$
$
\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 $n$
$
\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.
$
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 $n$
$
3n + 1 = 0 \\
n = - \dfrac{1}{3} \\
$
And
$
n - 4 = 0 \\
n = 4 \\
$
But, $n$ has to be a natural number.
Therefore, $n = 4$
Note: Many students make mistakes by using the formula of combination and not the formula of permutation. Here, in the expression, $^n{P_r}$, $P$ represents permutation and is equal to $\dfrac{{n!}}{{\left( {n - r} \right)!}}$. And, the formula of $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, where $C$ represents combination.
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