Question 557cf

Feb 11, 2015

n = 6 (which can be verified by substituting n=6 into each side of the equation).

P(2n+4,3) = (2/3)(P(n+4,4)#
by definition of permutation and multiplying both sides by 3:
$\left(3\right) \left(2 n + 4\right) \left(2 n + 3\right) \left(2 n + 2\right) = \left(2\right) \left(n + 4\right) \left(n + 3\right) \left(n + 2\right) \left(n + 1\right)$

factoring out 2's from two of the right side terms:
$\left(3\right) \left(2\right) \left(n + 2\right) \left(2 n + 3\right) \left(2\right) \left(n + 1\right) = \left(2\right) \left(n + 4\right) \left(n + 3\right) \left(n + 2\right) \left(n + 1\right)$

dividing out the (n+2) and (n+1) terms on both sides:
$\left(3\right) \left(2\right) \left(2 n + 3\right) \left(2\right) = \left(2\right) \left(n + 4\right) \left(n + 3\right)$

simplifying:
$24 n + 36 = 2 {n}^{2} + 14 n + 24$

into standard form:
$2 {n}^{2} - 10 n - 12 = 0$

using the standard formula for quadratic solution:
$\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

= $\frac{10 \pm \sqrt{100 + 96}}{4}$

= $\frac{10 \pm 14}{4}$

= $6$ or $- 1$ but $P \left(n + 4 , 4\right)$ is not defined for $n = - 1$

So, $n = 6$