# [✓] Solve the following inequality?

GROUPS:
 Here's the code: NSolve[(10!)/(10^n)*(1/(10 - n)!) < 0.5, n] but when I use the NSolve Command (as I want a numerical solution), I just get the following: NSolve[(567 2^(8 - n) 5^(2 - n))/(10 - n)! < 0.5, n] Please help me with this. Thanks in advance :)
10 months ago
2 Replies
 Marco Thiel 1 Vote Dear Raivat Shah,are you looking for a solution over the Reals or Integers? I guess that it is a good idea to start with the equalities such as: FindInstance[(10!)/(10^n)*(1/(10 - n)!) == 0.5 && n > 0, n, Reals] {{n -> 4.01848}} or else FindInstance[(10!)/(10^n)*(1/(10 - n)!) == 0.5 && n > 0, n, Integers] {} You might want to look at: Plot[{(10!)/(10^n)*(1/(10 - n)!), 0.5}, {n, 0, 20}, PlotRange -> All] and Plot[{(10!)/(10^n)*(1/(10 - n)!), 0.5}, {n, 0, 80}, PlotRange -> All] Also look at this Table[(10!)/(10^n)*(1/(10 - n)!), {n, 1, 1000}] and this (you might want to play with the PlotPoints option) LogPlot[Abs@{(10!)/(10^n)*(1/(10 - n)!), 1/2}, {n, 0, 60}, PlotRange -> All, PlotPoints -> 10000] Can you guess what is going on? Once you see that, it might help you to solve the original inequality.Best wishes,MarcoPS: This here might also be useful: Plot[n!, {n, -10, 0}]
10 months ago
 Dear Marco, Thank you so much. I was not only able to solve the inequality, but you have given me an excellent idea/way to present it graphically. I don't have enough words to thank you. I also realized that Mathematica is so powerful that it can do the computations for very large numbers. For example, FindInstance[(10000!)/(10000^n)*(1/(10000 - n)!) == 0.5 && n > 0, n, Reals] {{n -> 118.011}} Thank you once again, Raivat Shah