Krdytkyu

So going back to my old number theory book by I. Niven, An introduction to the theory on numbers, I saw these problems:

Question 1: Find the triplets of number $(x,y,z)$ such that $x+y+z=119,$ and none exceed $59?$

Question 2: Find four integers $(x,y,z,w)$, such that they have the property $xge 0 , yge 1, zge 2, wge 3$, and satisfies $x+y+z+w = 15.$

I would appreciate the indepth explanation to solving these problems

These are linear equations for which there is an established algorithm to solve, in terms of paramatrisations. I'll do the first question; see if you can generalise the idea to the second.
Recall that solving a linear equation in two variables over $mathbb Z$ is done like this: if $ax+by=c$, then find a particular solution $(x_0,y_0)$. Then all the solutions are given by $x=x_0-bt$, $y=y_o+at$, where $t$ ranges over all integral values. (Can you prove this?) Now let's move on to the three-variable case.

Set $u+v=k$. We have the simultaneous equations
$$begin{cases}u+v=k,\k+w=119.end{cases}$$
If we treat $k$ like a constant in the first equation we obtain the paramatrisation $u=k-t_1$, $v=t_1$. Treating $k$ like a variable in the second, we have $k=119-t_2$, $w=t_2$. Now, $u=k-t_1=119-t_1-t_2$. Hence, all the possible integer solutions are given by $(u,v,w)=(119-t_1-t_2,t_1,t_2)$. But none of the three variables exceed $59$, so we have the bounds $119-t_1-t_2leq59$, $t_1leq 59$, $t_2leq 59$. In other words, $t_1,t_2$ are any two integers so that each is not exceeding $59$ and their sum is at least $60$. This provides a complete paramatrisation and hence solution to the problem.

you can model such these equations like this:

first Q2)

suppose we have 15 identical balls and want to put them in 4 different baskets labeled as x, y, z, w such that basket y must has at least 1 ball inside it, basket z must has at least 2 balls inside it and basket w must has 3 balls inside it.

so we initially satisfy these criterias by just putting 1, 2 and 3 balls in baskets y, z and w respectively.
now we have 15-(1+2+3)=9 balls left with 3 basket which have no specific criterias: x + y' + z' + w' = 9

now if we put 3 delimiters between 9 balls to put them into 4 baskets equivalently: sections 1, 2, 3 and 4 corresponds to baskets x, y, z and w respectively.

one possible division is as follows: "{} | {OOO} | {} | {OOOOOO}" ("O" represents a ball and "|" represents a delimiter). its corresponding answer is (x, y', z', w') = (0, 3, 0, 6) that satisfies the equation x + y' + z' + w' = 9. so (x, y, z, w) = (0, 3+1, 0+2, 6+3) is a valid answer for main equation x + y + z + w = 15.

so the problem is to find all the linear permutations of 9 identical balls and 3 identical delimiters. and finally here is the number of valid (x, y, z, w) answers :

$frac{(9+3)!}{9!3!} = frac{{12}times{11}times{10}}{6} = 220$

Q1)

the problem is to find the number of (x, y, z) answers that satisfy the following equation:

$D sim [{x + y + z = 119; {0}leq{x}leq{59}, {0}leq{y}leq{59}, {0}leq{z}leq{59}}]$

now let:

$I sim [{x + y + z = 119; {0}leq{x}, {0}leq{y}, {0}leq{z}}]$
$I_1 sim [{x + y + z = 119; {60}leq{x}, {0}leq{y}, {0}leq{z}}]$
$I_2 sim [{x + y + z = 119; {0}leq{x}, {60}leq{y}, {0}leq{z}}]$
$I_3 sim [{x + y + z = 119; {0}leq{x}, {0}leq{y}, {60}leq{z}}]$

by inclusion-exclusion-identity we have:

$D = {bar{I_1}}cap{bar{I_2}}cap{bar{I_3}} = overline{({I_1}cup{I_2}cup{I_3})} = I - (I_1 + I_2 + I_3 - {I_1}
cap{I_2} - {I_1}cap{I_3} - {I_2}cap{I_3} + {I_1}cap{I_2}cap{I_3})$

also:

${I_1}cap{I_2} sim [{x + y + z = 119; {60}leq{x}, {60}leq{y}, {0}leq{z}}] rightarrow unsatisfiable$
${I_1}cap{I_3} sim [{x + y + z = 119; {60}leq{x}, {0}leq{y}, {60}leq{z}}] rightarrow unsatisfiable$
${I_2}cap{I_3} sim [{x + y + z = 119; {0}leq{x}, {60}leq{y}, {60}leq{z}}] rightarrow unsatisfiable$
${I_1}cap{I_2}cap{I_3} sim [{x + y + z = 119; {60}leq{x}, {60}leq{y}, {60}leq{z}}] rightarrow unsatisfiable$

so the answer is:

$D = frac{(119+2)!}{119!2!} - {3}times{frac{(119-60+2)!}{(119-60)!2!}} + {3}times{0} - 0 = frac{{121}times{120}}{2} - {3}times{frac{{61}times{60}}{2}} = 1770$