@Markiyan Hirnyk Maybe it's a PDF if x is restricted to x >= 0. Many (perhaps most) distributions have such a restriction.

@JohnS The scanf family of commands (fscanf, sscanf, and scanf) always return a list. Even if all the scans fail, they return the empty list; if you're scanning only one item, they return a list with that one item.

@Josolumoh Hold your horses. You've been very slow at providing the necessary details. Is it a continuous or discrete distribution? What is the support, i.e., the domain for x? What are the domains of the parameters b, d, and r?

The appropriate command is Statistics:-Sample.

@bloodtalon What is your Maple version? Like I said, it won't work with Maple earlier than 18, when the key option to sort was introduced. But Markiyan's adjustment (immediately above)  will make it work for earlier Maple.

@Josolumoh Is your function a probability density function (pdf)? If so, what's the name? If it's a standard one, Maple likely already has a sampling procedure for it.

@JohnS If L is a list or set, then L[] extracts the underlying sequence. This is documented in the ninth paragraph of Description at ?list, which simply states

To extract the contents of a list or set, use the empty selection operator [ ].

That's not possible: The function is much too complicated. Perhaps if you describe why you want to invert it, we could come up with an alternative.

You can only solve an equation for the highest derivative of each function. What is the reason that you want to solve for the second derivative of f4 rather than the third derivative?

What's sys? How's Maple supposed to know the ODEs that you've written on the paper just from sys? And I don't see anything about y(t) on your paper.

@Markiyan Hirnyk Yes, I like your way better. To handle more-general cases, change the key to (c-> evalf(rhs(c[-1]))).

@patient It can be done like this

linspace:= (a,b,n)-> a +~ (b-a)/(n-1)*[$0..n-1]:
U1:= eval(U(x), res):
U2:= proc(alpha)
local x;
     U1(parameters= [alpha]);
     plots:-spacecurve([x, alpha, U1(x)], x= -1..-.02, color= black, thickness= 0)
end proc:
plots:-display(
     [seq(U2(alpha), alpha= linspace(0.1, 10, 23))],
     axes= frame, labels= [x, alpha, ``], orientation= [-86, 68]
);

@patient Here's how to do the 3-D plot that you want:

U1:= eval(U(x), res):
U2:= proc(x,alpha)
     U1(parameters= [alpha]);
     U1(x)
end proc:
plot3d(U2, -1..-0.02, 0.1..10, labels= [x, alpha, U], axes= boxed);

@Earl Better yet, any polygon can be partitioned into triangles whose vertices are vertices of the polygon.

@Carl Love I made an incorrect generalization from the n=2 to n=3 case. At first I thought that no group of three should occur more than once. Actually, it is much more strict: No pair can appear as part of one of the groups of three more than once. This seems to be a much more difficult problem, and the number of solutions is far less than 51 or 55. Brian's generalization of Kitonum's solution only applies to the more-liberal case: that no group of three occurs more than once.

@ben2015 It's easy. Use a Vector, not a list. Use add and mul rather than sum and product.

A:= <a||(1..10)>:
F:= unapply(add(A(k)*(1-A(k))^2*mul((1-2*A(l))^3, l= 1..k-1), k= 1..10), a||(1..10));