The equation is given by:
3nP4=(n−1)P5
⇒(3n−4)!(3n)!=(n−6)!(n−1)![∵nPr=(n−r)!n!]
⇒(3n−4)!3n(3n−1)(3n−2)(3n−3)(3n−4)!= (n−6)!(n−1)(n−2)(n−3)(n−4)(n−5)(n−6)! [∵n!=n×(n−1)×(n−2).........4×3×2×1]
⇒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 ]
⇒(9n2−3n)(9n2−9n−6n+6)= (n2−2n−n+2)(n2−4n−3n+12)(n−5)
⇒(9n2−3n)(9n2−15n+6)= (n2−3n+2)(n2−7n+12)(n−5)
⇒(81n4−135n3+54n2−27n3+45n2−18n)= (n4−7n3+12n2−3n3+21n2 −36n+2n2−14n+24)(n−5)
⇒81n4−162n3+99n2−18n= (n4−10n3+35n2−50n+24)(n−5)
⇒81n4−162n3+99n2−18n= n5−5n4−10n4+50n3+35n3−175n2−50n2+250n+24n−120
⇒81n4−162n3+99n2−18n= n5−15n4+85n3−225n2+274n−120
⇒n5−15n4−81n4+85n3+162n3−225n2−99n2+274n+18n−120=0 [ Taking all the like terms to the left hand side of the equation ]
⇒n5−96n4+247n3−324n2+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≈1.04,n≈93,n≈0.28452+1.07508i andn≈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=1orn=1.04 [∵nPr=(n−r)!n!ifn>r]
Hence:
The value of nis93
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