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:

3nP4=(n1)P53n_{P_{4}} = (n-1)_{P_{5}}

(3n)!(3n4)!=(n1)!(n6)![nPr=n!(nr)!]\Rightarrow \frac{(3n)!}{(3n -4)!} = \frac{(n-1)!}{(n-6)!} \hspace{1 cm} \left[ \because n_{P_{r}} = \frac{n!}{(n-r)!} \right]


3n(3n1)(3n2)(3n3)(3n4)!(3n4)!=\Rightarrow \frac{3n(3n-1)(3n-2)(3n-3)(3n-4)!}{(3n-4)!} = (n1)(n2)(n3)(n4)(n5)(n6)!(n6)!\frac{(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)!}{(n-6)!} \hspace{1 cm} [n!=n×(n1)×(n2).........4×3×2×1]\left[ \because n! = n \times (n-1) \times (n-2) ......... 4 \times 3 \times 2 \times 1 \right]


3n(3n1)(3n2)(3n3)=\Rightarrow 3n(3n-1)(3n-2)(3n-3) = (n1)(n2)(n3)(n4)(n5)(n-1)(n-2)(n-3)(n-4)(n-5) [ By canceling (3n4)!(3n-4)! from both denominator and numerator of the left fraction and (n6)!(n-6)! from both numerator and denominator of the right fraction of the equation ]

(9n23n)(9n29n6n+6)=\Rightarrow (9n^2 - 3n)(9n^2 - 9n - 6n + 6) = (n22nn+2)(n24n3n+12)(n5)(n^2 -2n - n +2)(n^2 - 4n - 3n + 12)(n-5)

(9n23n)(9n215n+6)=\Rightarrow (9n^2 -3n)(9n^2- 15n + 6) = (n23n+2)(n27n+12)(n5)(n^2 - 3n + 2)(n^2 - 7n + 12)(n-5)

(81n4135n3+54n227n3+45n218n)=\Rightarrow (81n^4 - 135n^3 + 54n^2 - 27n^3 + 45n^2 - 18n) = (n47n3+12n23n3+21n2(n^4 - 7n^3 + 12n^2- 3n^3+ 21n^2 36n+2n214n+24)(n5)-36n + 2n^2 - 14n + 24)(n-5)

81n4162n3+99n218n=\Rightarrow 81n^4 - 162n^3 + 99n^2 - 18n = (n410n3+35n250n+24)(n5)(n^4 - 10n^3 + 35n^2 - 50n + 24)(n-5)

81n4162n3+99n218n=\Rightarrow 81n^4 - 162n^3 + 99n^2 - 18n = n55n410n4+50n3+35n3175n250n2+250n+24n120n^5 - 5n^4 - 10n^4 + 50n^3 + 35n^3 - 175n^2 - 50n^2 + 250n + 24n - 120

81n4162n3+99n218n=\Rightarrow 81n^4 - 162n^3 + 99n^2 - 18n = n515n4+85n3225n2+274n120n^5 -15n^4 + 85n^3 - 225n^2 + 274n - 120

n515n481n4+85n3+162n3225n299n2+274n+18n120=0\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 ]

n596n4+247n3324n2+292n120=0\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=1n = 1 ,n1.04,n93,n0.28452+1.07508i, \,\, n \approx 1.04 \,\,, \,\, n \approx 93 \,\, , \,\, n \approx 0.28452 + 1.07508i \,\, andn0.284521.07508iand \,\, n \approx 0.28452 - 1.07508i

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

The two imaginary values of nn where ii included, we ignore it. Because nn always takes real values.

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

Hence:

The value of nis93n \,\, is \,\, 93


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