Discrete Mathematics – Total Number of Integral Solutions with Constraints

Find the number of ways in which 5 dice can be rolled to get 25.

While solving this question, the way we solve it is $$x_1 + x_2 + x_3 + x_4 + x_5$$ $$= 25$$ or $$1 <= x_i <= 6$$

So we replace $$x_i$$ by $$y_i = 6-x_i$$ , Which one is $$x_i = 6-y_i$$

replacing $$x_i$$ in the equation above, we get it like → $$(6 * 5) – (y_1 + y_2 + y_3 + y_4 + y_5)$$ $$= 25$$

$$(y_1 + y_2 + y_3 + y_4 + y_5)$$ = $$5$$

After solving this equation with the solution solution $$(n-r + 1)! / (n! * (r-1)!)$$ we get years like → $$126$$

Now consider this problem,

The number of non-negative whole solutions such as $$x_1 + x_2 + x_3 = 17$$ or $$x_1> 1, x_2> 2, x_3> 3$$ is ___________________

Solving this, we solve it as → $$y_1 = x_1-2$$ , $$y_2 = x_2 -3$$ , $$y_3 = x_3-4$$

so, $$x_1 = y_1 + 2$$ , $$x_2 = y_2 + 3$$ , $$x_3 = y_3 + 4$$

Now, we substitute this in our original equation to get →

$$y_1 + 2 + y_2 + 3 + y_3 + 4 = 17$$

$$y_1 + y_2 + y_3 = 8$$

and after solving that, we get the years as $$45$$

Now I have a $$DOUB$$ here in the second problem since when $$x_1> 1$$ we do it like $$x_1 = y_1 + 2$$ but in the first problem, all the dice should have a value $$> 0$$ , so why in this case we did not do $$x_i = y_i + 1$$ for all cases?

And elsewhere if the question was like

$$x_1 + x_2 + x_3 = 12$$ , $$2 <= x <= 5$$ so how to solve this problem by using an entire solution and applying the formula $$(n-r + 1)! / (n! * (r-1)!)$$ ?

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Equation Resolution – Defines a function that includes two resolution solutions

I'm trying to combine the solutions from both NSolve and define a new function that I have to integrate.

Basically, I defined two NSolve with the names eqpart and usol. I've assigned their solutions to three functions, xtot (Tx), ne (y) and T (y).

Then I defined neff (y_): = ne (y) * xtot (T (y)) and I try to integrate it as neff0 = NIntegrate (neff (y) * Lc * 2 * Pi * Rc ^ 2 * y, {y, ymin, ymax}) / (Pi *
Rc ^ 2 * Lc);

As an output, I have several errors saying that "eqpart … is neither a list of replacement rules, nor a valid dispatch table, and therefore can not be used to replace"

What am I doing wrong?

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Handling list – correctly evaluate the existence of great solutions

I'm trying to evaluate correctly

$$D (S cap (a, b)) = lim_ {n to infty} frac { left | S cap {F_n cap (a, b)} right |} { left | F_n cap (a, b) right |$$

or $$D$$ is the density of $$S cap (a, b)$$ (in $$A cap (a, b)$$) $$(a, b)$$ is an interval for $$a, b in mathbb {R}$$, $$F_n$$ is the Folner sequence of $$A$$, and $$S subseteq A$$. For more information, click here (replace $$G, X, i, g$$ with $$A, S, n, a$$), click (here) and here.

I put $$A = left { frac {p} {2 ^ k (2q + 1)}: p, q, k in mathbb {Z}, k ge 0 right$$, $$F_n = left { frac {p} {2 ^ k (2q + 1)}: p, q, k in mathbb {Z}, k n, | 2q + 1 | n right }$$ and $$S = left { frac {j ^ 2} {k ^ 3}: j, k in mathbb {Z}, k neq 0 right$$

In my code $$A$$ is A(p_,k_,q_), $$F_n$$ is f(n_,a_,b_), $$S$$ is S(j_,k_) and $$D$$ is d.

Clear(A, F, f, p, Ff, S, X, Y, d, j, k, q, n, a, b);
A(p_, k_, q_) := p/((2^k)*(2*q + 1));
F(p_, n_) :=
Table(A(p, k, q), {k, 0, Floor(Log(2, n))}, {q, 0,
Floor((n - 1)/2)});
f(n_, a_, b_) :=
p /. Table(
Solve(a <= A(p, k, q) <= b, p, Integers), {k, 0,
Floor(Log(2, n))}, {q, 0, Floor((n - 1)/2)});
Ff(n_, a_, b_) :=
DeleteDuplicates@
Flatten@Table(
F(f(n, a, b)((v))((u)), n)((v))((u)), {v, 1,
Floor(Log(2, n)) + 1}, {u, 1, Floor((n - 1)/2) + 1});
RandomsampleFf(n_, a_, b_) := RandomSample(Ff(n, a, b), 100)
S(j_, k_) := (j^2)/(k^3);
X(n_, a_, b_) :=
Count(Resolve(
Exists({j, k}, S(j, k) == # && {j, k} (Element) Integers)) & /@
RandomsampleFf(n, a, b), True);
Y(n_, a_, b_) := Length(RandomsampleFf(n, a, b));
d(n_, a_, b_) := N((X(n, a, b))/Y(n, a, b));
RandomsampleFf(40, 1, 2)
Num = X(40, 1, 2)
Den = Y(40, 1, 2)
Timing(N(Num/Den))

My result is

{2329/1184, 1571/992, 719/432, 783/608, 325/272, 557/304, 1375/992,
403/240, 1779/1120, 1767/1184, 2355/1184, 103/72, 1821/992, 867/800,
505/464, 581/544, 365/312, 23/12, 947/608, 1735/1184, 1907/1120,
335/288, 229/184, 1243/736, 1163/672, 67/50, 1487/1248, 377/280,
15/8, 1999/1184, 379/200, 809/416, 859/496, 1569/1120, 569/544,
1841/928, 1469/864, 101/88, 355/336, 631/416, 907/480, 607/312,
341/296, 66/35, 531/352, 257/160, 619/400, 467/248, 1659/928,
213/200, 573/560, 283/224, 493/336, 273/232, 1873/1248, 305/216,
1005/928, 1059/992, 597/352, 61/38, 137/76, 203/132, 825/736,
761/736, 57/52, 583/432, 1123/624, 195/124, 779/464, 223/140,
1611/1120, 939/800, 261/248, 891/736, 329/232, 1343/1184, 2073/1184,
225/208, 785/608, 521/432, 707/464, 477/248, 38/23, 543/272, 171/104,
519/280, 1071/800, 1487/800, 557/352, 811/528, 677/432, 85/48,
1295/1248, 379/224, 1559/928, 1483/864, 72/37, 403/216, 71/36, 231/136}
4
100
{0., 0.04}

The density is 0.04 instead of 1. This is due to the fact Exists incorrectly evaluates higher values ​​of RandomSampleFf(n,a,b). How do we correct Exists if the values ​​of RandomSampleFf(n,a,b) are arbitrary.

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permutations – Number of solutions and minimum cues in Sixy Sudoku

Sixy Sudoku is a variation on the Latin places and the traditional sudoku played on a $$6 times 6$$ grid with an initial index of several cells filled with a subset of digits $$1$$$$6$$. The task is to fill the remaining cells so that each number appears once in each

• $$1 times 6$$ row
• $$6 times 1$$ column
• $$2 times 3$$ shaded rectangle
• $$3 times 2$$ delimited rectangle Questions

• Given a grid without initial filled cells, how many valid filled grids, $$K$$, is there (up to the symmetry of permutation of numbers)?
• What is the minimum number of cells filled, $$n ^ *$$, which guarantees a unique solution?
• For this minimum $$n ^ *$$, how many distinct cell locations guarantee a unique solution (up to the permutation symmetry of digits)?

For the first problem, without loss of generality, we can define the numbers in the shaded rectangle at the top left, as shown below: A naive upper limit on the number of valid filled grids, $$K$$consists in considering each of the remaining shaded rectangles as independent, which gives $$(6!) ^ 5$$ solutions. (The same logic applies to the consideration of independent lines, or independent columns, or separate rectangles.) Of course, this limit will be extremely loose because it will not incorporate many constraints.

A slightly narrower limit can be found by considering the left set of three shaded rectangles as independent, and then adding the row constraint for each shaded rectangle aligned to the left. That way we get $$(6!) ^ 2 ((3!) ^ 2) ^ 3$$. But of course, this bound does not include any constraints, such as column constraint. A narrow limit on the number of bits needed to specify a problem (initial cell specifications) with a single solution is $$log_2 K$$.

For the last two problems, it will be interesting to see the proximity of the information defined by the minimum number of cells filled, $$n ^ *$$ (or $$n ^ * geq 5$$ for the specification of figures), and the candidate placements approximate the related information given by $$K$$.

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Equation in Resolution – Reduce and Expand Solutions After Resolution

I solve a system of 4 equations with 4 variables z_i (see below).

I would like to change the way the results are written as such for each i:
y_i = AX + sum (By _ (- i)) + Cz_i + u_i

where the sum comes from other indices that i, u_i groups all the terms other than X, y _ (- i) and z_i.

for example:
y_1 = AX + By_2 + Cy_3 + Dy_4 + Ez_1 + u_1

Is it possible to do it?

Thank you.

Solve({Subscript((Pi), 11) Subscript(z, 1) +
Subscript((Pi), 12) Subscript(z, 2) +
Subscript((Pi), 13) Subscript(z, 3) +
Subscript((Pi), 14) Subscript(z, 4) + Subscript((Pi), 10) X +
Subscript((Nu), 1) == Subscript(y, 1),
Subscript((Pi), 21) Subscript(z, 1) +
Subscript((Pi), 22) Subscript(z, 2) +
Subscript((Pi), 23) Subscript(z, 3) +
Subscript((Pi), 24) Subscript(z, 4) + Subscript((Pi), 20) X +
Subscript((Nu), 2) == Subscript(y, 2),
Subscript((Pi), 31) Subscript(z, 1) +
Subscript((Pi), 32) Subscript(z, 2) +
Subscript((Pi), 33) Subscript(z, 3) +
Subscript((Pi), 34) Subscript(z, 4) + Subscript((Pi), 30) X +
Subscript((Nu), 3) == Subscript(y, 3),
Subscript((Pi), 41) Subscript(z, 1) +
Subscript((Pi), 42) Subscript(z, 2) +
Subscript((Pi), 43) Subscript(z, 3) +
Subscript((Pi), 44) Subscript(z, 4) + Subscript((Pi), 40) X +
Subscript((Nu), 4) == Subscript(y, 4)}, {Subscript(z, 1),
Subscript(z, 2), Subscript(z, 3), Subscript(z, 4)})

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