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# If ${}^{{\text{n - 1}}}{{\text{P}}_{\text{3}}}{\text{:}}{}^{{\text{n + 1}}}{{\text{P}}_{\text{3}}}{\text{ = 5:12}}$, find n.

Last updated date: 20th Jun 2024
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Hint: We’ll approach the value of n by simplifying the equation ${}^{{\text{n - 1}}}{{\text{P}}_{\text{3}}}{\text{:}}{}^{{\text{n + 1}}}{{\text{P}}_{\text{3}}}{\text{ = 5:12}}$, for the simplification of this equation we’ll use of the formula
${}^{\text{n}}{{\text{P}}_{\text{r}}}{\text{ = }}\dfrac{{{\text{n!}}}}{{\left( {{\text{n - r}}} \right){\text{!}}}}$
Further, we’ll obtain a quadratic equation in ‘n’ then on solving that quadratic equation for ‘n’ we’ll get two values for it as ‘n’ is a natural number, we’ll get our answer.

Complete step by step answer:

Given data: ${}^{{\text{n - 1}}}{{\text{P}}_{\text{3}}}{\text{:}}{}^{{\text{n + 1}}}{{\text{P}}_{\text{3}}}{\text{ = 5:12}}$
Now, solving for ${}^{{\text{n - 1}}}{{\text{P}}_{\text{3}}}{\text{:}}{}^{{\text{n + 1}}}{{\text{P}}_{\text{3}}}{\text{ = 5:12}}$
$\Rightarrow \dfrac{{{}^{{\text{n - 1}}}{{\text{P}}_{\text{3}}}}}{{{}^{{\text{n + 1}}}{{\text{P}}_{\text{3}}}}}{\text{ = }}\dfrac{{\text{5}}}{{{\text{12}}}}$
Using, ${}^{\text{n}}{{\text{P}}_{\text{r}}}{\text{ = }}\dfrac{{{\text{n!}}}}{{\left( {{\text{n - r}}} \right){\text{!}}}}$
$\dfrac{{\dfrac{{{\text{(n - 1)!}}}}{{\left( {{\text{n - 1 - 3}}} \right){\text{!}}}}}}{{\dfrac{{{\text{(n + 1)!}}}}{{\left( {{\text{n + 1 - 3}}} \right){\text{!}}}}}}{\text{ = }}\dfrac{{\text{5}}}{{{\text{12}}}}$
$\Rightarrow \dfrac{{{\text{(n - 1)!}}}}{{\left( {{\text{n - 1 - 3}}} \right){\text{!}}}}\dfrac{{\left( {{\text{n + 1 - 3}}} \right){\text{!}}}}{{{\text{(n + 1)!}}}}{\text{ = }}\dfrac{{\text{5}}}{{{\text{12}}}}$
On simplification we get,
$\Rightarrow \dfrac{{{\text{(n - 1)!}}}}{{\left( {{\text{n - 4}}} \right){\text{!}}}}\dfrac{{\left( {{\text{n - 2}}} \right){\text{!}}}}{{{\text{(n + 1)!}}}}{\text{ = }}\dfrac{{\text{5}}}{{{\text{12}}}}$
On Using ${\text{n! = n(n - 1)!}}$, we get,
$\dfrac{{{\text{(n - 1)!}}}}{{\left( {{\text{n - 4}}} \right){\text{!}}}}\dfrac{{\left( {{\text{n - 2}}} \right){\text{(n - 3)(n - 4)!}}}}{{{\text{(n + 1)n(n - 1)!}}}}{\text{ = }}\dfrac{{\text{5}}}{{{\text{12}}}}$
On cancelling common terms we get,
$\Rightarrow \dfrac{{\left( {{\text{n - 2}}} \right){\text{(n - 3)}}}}{{{\text{(n + 1)n}}}}{\text{ = }}\dfrac{{\text{5}}}{{{\text{12}}}}$
After cross multiplication we get,
$\Rightarrow {\text{12}}\left( {{\text{n - 2}}} \right){\text{(n - 3) = 5(n + 1)n}}$
On expansion we get,
$\Rightarrow {\text{12(}}{{\text{n}}^{\text{2}}}{\text{ - 3n - 2n + 6) = 5(}}{{\text{n}}^{\text{2}}}{\text{ + n)}}$
$\Rightarrow {\text{12(}}{{\text{n}}^{\text{2}}}{\text{ - 5n + 6) = 5(}}{{\text{n}}^{\text{2}}}{\text{ + n)}}$
On further simplification we get,
$\Rightarrow {\text{12}}{{\text{n}}^{\text{2}}}{\text{ - 60n + 72 = 5}}{{\text{n}}^{\text{2}}}{\text{ + 5n}}$
$\Rightarrow {\text{12}}{{\text{n}}^{\text{2}}}{\text{ - 5}}{{\text{n}}^{\text{2}}}{\text{ - 60n - 5n + 72 = 0}}$
$\Rightarrow {\text{7}}{{\text{n}}^{\text{2}}}{\text{ - 65n + 72 = 0}}$
Now we’ll split the coefficient of n such that they are the factors of the coefficient of ${{\text{n}}^{\text{2}}}$ and independent term, we get,
$\Rightarrow {\text{7}}{{\text{n}}^{\text{2}}}{\text{ - (56 + 9)n + 72 = 0}}$
$\Rightarrow {\text{7}}{{\text{n}}^{\text{2}}}{\text{ - 56n - 9n + 72 = 0}}$
On taking factors common we get,
$\Rightarrow {\text{7n(n - 8) - 9(n - 8) = 0}}$
After taking (n-8) common from both the terms we get,
$\Rightarrow {\text{(n - 8)(7n - 9) = 0}}$
${\text{i}}{\text{.e n - 8 = 0 or 7n - 9 = 0}}$
$\therefore {\text{n = 8 or n = }}\dfrac{{\text{9}}}{{\text{7}}}$
Since n is a natural number
Note: A permutation is selecting all the ordered pair of ‘r’ elements out of ‘n’ total elements is given by ${}^{\text{n}}{{\text{P}}_{\text{r}}}$, and this expression is equal to
${}^{\text{n}}{{\text{P}}_{\text{r}}}{\text{ = }}\dfrac{{{\text{n!}}}}{{\left( {{\text{n - r}}} \right){\text{!}}}}$
It can also be said for arranging all the elements in order after selecting combinations of ‘r’ element out of total ‘n’ elements, where expression for combination is ${}^{\text{n}}{{\text{C}}_{\text{r}}}$, and this expression is equal to
${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
${}^{\text{n}}{{\text{P}}_{\text{r}}}{\text{ = r!}}{}^{\text{n}}{{\text{C}}_{\text{r}}}$
$\begin{gathered} {}^{\text{n}}{{\text{P}}_{\text{r}}}{\text{ = r!}}\dfrac{{n!}}{{r!\left( {n - r} \right)!}} \\ \Rightarrow {}^{\text{n}}{{\text{P}}_{\text{r}}}{\text{ = }}\dfrac{{n!}}{{\left( {n - r} \right)!}} \\ \end{gathered}$