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

Question #102920
Find the value of n given that 3n permutation 4 = (n-1)permutation 5
1
Expert's answer
2020-02-23T17:02:16-0500

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"


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS