Question
If P(n ,5) = 20.P(n, 3), find n.
P(n ,5) = 20.P(n, 3)
$\Rightarrow \frac{\text{n!}}{(\text{n-5})!}=20\times\frac{\text{n!}}{(\text{n-3)}!}$ $\Rightarrow \frac{1}{(\text{n}-5)}!= \frac{20}{(\text{n}-3)(\text{n}-3 -1)(\text{n}-3-2)!}$ $\Rightarrow \frac{1}{(\text{n}-5)!}= \frac{20}{(\text{n}-3)(\text{n}-4)\text{n}-5)!}$ $\Rightarrow \frac{(\text{n}-3)(\text{n}-4)(\text{n}-5)!}{(\text{n}-5)!}= 20$ $\Rightarrow (\text{n}-3)(\text{n}-4) =20$ $\Rightarrow \text{n}^2-4\text{n}-3\text{n}-8=0$ $\Rightarrow \text{n}^2+7\text{n}-8=0$ $\Rightarrow \text{n}^2-8\text{n}+1\text{n}-8=0$ $\Rightarrow \text{n}(\text{n}-8)+1(\text{n}-8)= 0$ $\Rightarrow (\text{n}-8)(\text{n}+1)= 0$ $\Rightarrow \text{n}-3 = 0$ $\Rightarrow \text{n} = 8 [\because\text{n}\neq-1]$ Hence, $ \text{n} = 8$Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\frac{1}{3}+\frac{1}{5^2}+\frac{1}{3^3}+\frac{1}{5^4}+\frac{1}{3^5}+\frac{1}{5^6}+\ ...\infty$