Answer to Question #102920 in Combinatorics | Number Theory for Glory okoli

Question

Find the value of n given that 3n permutation 4 = (n-1)permutation 5

Answer 1

The equation is given by:

$3n_{P_{4}} = (n-1)_{P_{5}}$

$\Rightarrow \frac{(3n)!}{(3n -4)!} = \frac{(n-1)!}{(n-6)!} \hspace{1 cm} \left[ \because n_{P_{r}} = \frac{n!}{(n-r)!} \right]$

$\Rightarrow \frac{3n(3n-1)(3n-2)(3n-3)(3n-4)!}{(3n-4)!} =$ $\frac{(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)!}{(n-6)!} \hspace{1 cm}$ $\left[ \because n! = n \times (n-1) \times (n-2) ......... 4 \times 3 \times 2 \times 1 \right]$

$\Rightarrow 3n(3n-1)(3n-2)(3n-3) =$ $(n-1)(n-2)(n-3)(n-4)(n-5)$ [ By canceling $(3n-4)!$ from both denominator and numerator of the left fraction and $(n-6)!$ from both numerator and denominator of the right fraction of the equation ]

$\Rightarrow (9n^2 - 3n)(9n^2 - 9n - 6n + 6) =$ $(n^2 -2n - n +2)(n^2 - 4n - 3n + 12)(n-5)$

$\Rightarrow (9n^2 -3n)(9n^2- 15n + 6) =$ $(n^2 - 3n + 2)(n^2 - 7n + 12)(n-5)$

$\Rightarrow (81n^4 - 135n^3 + 54n^2 - 27n^3 + 45n^2 - 18n) =$ $(n^4 - 7n^3 + 12n^2- 3n^3+ 21n^2$ $-36n + 2n^2 - 14n + 24)(n-5)$

$\Rightarrow 81n^4 - 162n^3 + 99n^2 - 18n =$ $(n^4 - 10n^3 + 35n^2 - 50n + 24)(n-5)$

$\Rightarrow 81n^4 - 162n^3 + 99n^2 - 18n =$ $n^5 - 5n^4 - 10n^4 + 50n^3 + 35n^3 - 175n^2 - 50n^2 + 250n + 24n - 120$

$\Rightarrow 81n^4 - 162n^3 + 99n^2 - 18n =$ $n^5 -15n^4 + 85n^3 - 225n^2 + 274n - 120$

$\Rightarrow n^5 - 15n^4 - 81n^4 + 85n^3 + 162n^3 - 225n^2 - 99n^2 + 274n + 18n -120 = 0$ [ Taking all the like terms to the left hand side of the equation ]

$\Rightarrow n^5 - 96n^4 + 247n^3 - 324n^2 + 292n - 120 = 0$

Now:

We can see that normally we can not find the solution of this equation. If we use graphing calculator then we will get the following roots of the equation:

$n = 1$ $, \,\, n \approx 1.04 \,\,, \,\, n \approx 93 \,\, , \,\, n \approx 0.28452 + 1.07508i \,\,$ $and \,\, n \approx 0.28452 - 1.07508i$

From these values of $n,$ we can say that:

The two imaginary values of $n$ where $i$ included, we ignore it. Because $n$ always takes real values.

We can not assign $n = 1 \,\, or \,\, n = 1.04$ $\left[ \because n_{P_{r}} = \frac{n!}{(n-r)!} \, \, if \,\, n > r \right]$

Hence:

The value of $n \,\, is \,\, 93$