Absolutely lazy mistake

The equality will be satisfied if and only if $a \ge b \ge c$ or $a \le b \le c$ . Thus, we want to count the number of monotone ordered triples taken from $[0, n-1]^3$ . In case of strict inequality, there are simply $2 \cdot \binom{n}{3}$ such triples. In the case where $a = b$ or $b = c$ but not both, there are $2 \cdot 2 \cdot \binom{n}{2}$ such triples (one factor of two accounts for nonincreasing versus nondecreasing triples, the other accounts for the two cases of the first pair being equal versus the second pair). Finally, in the case where $a = b = c$ , there are simply $\binom{n}{1} = n$ triples. Therefore, the total number of desired triples is $\binom{n}{1} + 4 \binom{n}{2} + 2 \binom{n}{3} = \frac{n}{3}(n^2+3n-1),$ and for $n = 11$ , this gives the value $\boxed{561}$ .

There are two cases: All of the absolute value expressions are non-negative, or all of them are non-positive. This yields $a \geq b \geq c$ OR $a \leq b \leq c$ .

In total, there are $11^3 = 1331$ triplets. Of these, $11$ are composed of three of the same integer, $11 * 10 * 9 = 990$ are composed of three distinct integers, and thus that leaves $330$ sets that are composed of two elements that are the same and one distinct. If we were to disregard order, this gives $11, 165, 110$ sets. With these un-ordered sets, a non-decreasing configuration and a non-increasing configuration each corresponds to one solution of the equation.

So, for the $11$ sets that are the same element, the non-decreasing and non-increasing configurations are the same , so there are $11$ ordered solutions. For the other two types of sets, they can be ordered in distinct non-decreasing and non-increasing formations, so each set corresponds to two ordered solution sets. This gives $2 \times 275 = 550$ . Thus, our answer is $\boxed{561}$ .

Alternative solution

Let's consider each inequality separately:

When $a \geq b \geq c$ , let's try to calculate this. When $a = 0$ then the solution is $(0, 0, 0)$ , when $a = 1$ the solutions are $(1, 0, 0) (1, 1, 0) (1, 1, 1) \rightarrow 1 + 2 = 3$ , when $a = 2$ the solutions are $(2, 0, 0) (2, 1, 0) (2, 1, 1) (2, 2, 0) (2, 2, 1) (2, 2, 2) \rightarrow 1 + 2 + 3 = 6$ . Clearly, these each correspond to a triangle number, so the total number would be equal to the sum of the first 11 triangle numbers, also known as the 11th tetrahedral number, which is $286$ .

When $a \leq b \leq c$ , clearly we get the same number of solutions. However, we double-count when $a = b = c$ , and this occurs $11$ times.

In summary, the total number of solutions would then be $286 \times 2 - 11 = \boxed{561}$ .

We do cases on the order of $a$ , $b$ , and $c$ .

If $a\ge b\ge c$ , $|a-c|=a-c=a-b+b-c=|a-b|+|b-c|$ so $a\ge b\ge c$ works.

If $c\ge b\ge a$ , $|a-c|=c-a=c-b+b-a=|a-b|+|b-c|$ so $c\ge b\ge a$ works.

If $a\ge c\ge b$ , $|a-c|=a-c$ $|a-b|+|b-c|=a-b+c-b=a+c-2b$ so if the two are equal, $a-c=a+c-2b\implies b=c$ so $a\ge c\ge b$ works if $b=c$ .

If $c\ge a\ge b$ , $|a-c|=c-a$ $|a-b|+|b-c|=a-b+c-b=a+c-2b$ so if the two are equal, $c-a=a+c-2b\implies a=b$ so $c\ge a\ge b$ works if $a=b$ .

If $b\ge a\ge c$ , $|a-c|=a-c$ $|a-b|+|b-c|=b-a+b-c=2b-a-c$ so if the two are equal, $a-c=2b-a-c\implies b=c$ so $b\ge a\ge c$ works if $a=b$ .

If $b\ge c\ge a$ , $|a-c|=c-a$ $|a-b|+|b-c|=b-a+b-c=2b-a-c$ so if the two are equal, $c-a=2b-a-c\implies b=c$ so $b\ge c\ge a$ works if $b=c$ .

Then, we see that $a=b$ always works, since it works if $a=b\ge c$ or $c\ge a=b$ from cases 1 and 4. Similarly, $b=c$ always works. The number of ways this can happen is $121+121-11=231$ , since we overcount the 11 when $a=b=c$ .

Also, we see that if $a\ge b\ge c$ or $c\ge b\ge a$ , it works. Since we already accounted for when some of the variables are equal, we can only consider when $a>b>c$ or $c>b>a$ . These two are symmetric, and for each, there are $\dbinom{11}{3}=165$ ways to pick $a$ , $b$ , and $c$ because each set of three numbers can be assigned in order.

The answer is $231+165+165=\boxed{561}$ .

Motivation: Absolute values are tricky, and I tried to remove them by checking all possible orderings. From there, the counting was a little tricky, but it worked. I am interested in seeing a simpler solution, if there is one.

The key is to realize that b is between a and c, as if for example b>a>c, then |a-b|+|b-c| = b-a+b-c; the b's will have the same sign and will not "cancel." Note that b can be equal to a or c. Now we just have to count all the ways that b can be equal to or between a and c.

Suppose the three are all distinct. Then there are 11 choose 3 ways to choose a triple. The middle number is b, but we can assign a and c in 2 ways, so we multiply by 2. (11 choose 3)*2 = 330 ways.

If b equals one of distinct a and c, then we must have 11 choose 2 pairs, times 2 for choosing which number b is equal to, times 2 again for choice of a and c. (11 choose 2) 2 2 = 220 ways.

Finally, if a, b, and c are all equal there are 11 possibilities. This leads to a grand total of 330+220+11=561 triples.

$|a-c|=a-c$ if and only if $a\ge c$ , and $a-c=|a-b|+|b-c|$ if and only if $a\ge b$ and $b\ge c$ .

$|a-c|=c-a$ if and only if $c\ge a$ , and $c-a=|a-b|+|b-c|$ if and only if $a\le b$ and $b\le c$ .

So we need to find all ordered triples of integers between $0$ and $10$ inclusive which satisfy $a\ge b\ge c$ or $a\le b\le c$ . For any given $b$ , there are $(b+1)$ ordered pairs $(b,c)$ satisfying $b\ge c$ as $0\le c\le b$ , so for any given $a$ there are $1 + 2 + 3 + \cdots+(a+1) = \frac{(a+1)(a+2)}{2}$ ordered triples $(a,b,c)$ satisfying $a\ge b\ge c$ . The total number of triples satisfying $a\ge b\ge c$ is thus the sum of the triangle numbers from $a=0$ to $a=10$ , which can be quickly calculated by substituting $x=11$ into the formula for the sum of the first $n$ triangle numbers: $\frac{x}{6}(x+1)(x+2)$ , which gives $286$ . Clearly the number of ordered triples satisfying $a\ge b\ge c$ is the same as the number of ordered triples satisfying $a\le b\le c$ , so we have $2\times 286 = 572$ . However we have counted triples where $a=b=c=$ twice, so we must subtract $11$ , giving us $561$ triples in total.

For the given condition to hold either a≤b≤c or a≥b≥c has to be true. Now we fix b from 0-10 and get the number of cases which does not comply with above conditions as follows:

For b=0, if a is 0 then we have no problem, but if a = 1 then we have problem if c = 1 to 10 so 10 cases and again a can take 10 values hence total is 10*10 = 100 cases

Similarly for b = 1 we have 1 case for a = 0 and 9*9=81 cases for a = 2,3,4,5,6,7,8,9,10.

Continuing we have 2 2+8 8 cases for b = 2, and we have 3 3+7 7 cases for b = 3, and we have 4 4+6 6 cases for b = 4, and we have 5 5+5 5 cases for b = 5, and we have 6 6+4 4 cases for b = 6, and we have 7 7+3 3 cases for b = 7, and we have 8 8+2 2 cases for b = 8, and we have 9 9+1 1 cases for b = 9, and we have 10 10 cases for b = 10, hence total number cases not complying with above conditions are 2 (1+4+9+16+25+36+49+64+81+100) = 770

Now total cases is 11 11 11 therefore cases where above condition holds are 1331- 770 = 561