Question

Find the value of n by solving the equation $\dfrac{{^{\text{n}}{{\text{P}}_{\text{4}}}}}{{^{{\text{n - 1}}}{{\text{P}}_{\text{4}}}}}{\text{ = }}\dfrac{{\text{5}}}{{\text{3}}}$ where, n>4.

Answer 1

Hint: In order to solve this problem you need to know the formula $^{\text{a}}{{\text{P}}_{\text{b}}}{\text{ = }}\dfrac{{{\text{a!}}}}{{{\text{(a - b)!}}}}$. Using this formula you will get the right answer to this question.

Complete step-by-step answer:
We need to find the value of n.
We have the equation $\dfrac{{^{\text{n}}{{\text{P}}_{\text{4}}}}}{{^{{\text{n - 1}}}{{\text{P}}_{\text{4}}}}}{\text{ = }}\dfrac{{\text{5}}}{{\text{3}}}$.
We know the formula $^{\text{a}}{{\text{P}}_{\text{b}}}{\text{ = }}\dfrac{{{\text{a!}}}}{{{\text{(a - b)!}}}}$………….(1)
So, we will expand only LHS of the equation $\dfrac{{^{\text{n}}{{\text{P}}_{\text{4}}}}}{{^{{\text{n - 1}}}{{\text{P}}_{\text{4}}}}}{\text{ = }}\dfrac{{\text{5}}}{{\text{3}}}$ with the help of (1).
So, $\dfrac{{^{\text{n}}{{\text{P}}_{\text{4}}}}}{{^{{\text{n - 1}}}{{\text{P}}_{\text{4}}}}} = \dfrac{{\dfrac{{n!}}{{(n - 4)!}}}}{{\dfrac{{(n - 1)!}}{{(n - 1 - 4)!}}}} = \dfrac{{n!\, \times \,(n - 5)!}}{{(n - 4)!\, \times \,(n - 1)!}} = \dfrac{n}{{n - 4}}\,\,\,\,\,\,.....................(2)$
From (1) and (2) we can do,
$
  \dfrac{{\text{n}}}{{{\text{n - 4}}}}{\text{ = }}\dfrac{{\text{5}}}{{\text{3}}} \\
  {\text{3n = 5n - 20}} \\
  {\text{2n = 20}} \\
  {\text{n = 10}} \\
$
Hence the value of n is 10.

Note: In this type of questions students may confuse with the formulas of $^{\text{a}}{{\text{P}}_{\text{b}}}$ and $^{\text{n}}{{\text{C}}_{\text{r}}}$.