Animated Solution for Mathematics - Permutations and Combinations: Let $P_1,P_2,\dots ,P_{15}$ be 15 points on a circle. The number of distinct triangles formed by points $P_i,P_j,P_k$ such that $i+j+k\ne 15$, is :

Imagine you are standing before a circle, and on its circumference, fifteen distinct points are marked, labeled $P_1$ through $P_{15}$. These points are the vertices of potential triangles.

In the world of geometry, any three points on a circle are inherently non-collinear. This is a beautiful, simplifying truth. It means that every single combination of three points you choose will inevitably form a triangle.

Our mission is to find how many of these triangles exist such that the sum of their indices, $i+j+k$, is not equal to $15$. We will employ the Complement Method to solve this.

Instead of hunting for the 'good' triangles, we will first embrace the entire universe of possibilities and then subtract the 'forbidden' ones.

To find the total number of triangles, we need to select any three points out of fifteen. This is a classic combination problem.

We use the formula:

Substituting $n=15$ and $r=3$, we get:

Calculating this, we find $5\times 7\times 13=455$. This is our total universe: $455$ possible triangles.

Now, we must identify the triangles that violate our condition. We are looking for triplets $(i,j,k)$ such that $1\le i<j<k\le 15$ and $i+j+k=15$.

This is a partition problem. We must be systematic by fixing $i$ and finding the pairs $(j,k)$ that satisfy $j+k=15-i$.

For $i=1$, $j+k=14$. Since $j>1$, $j$ can be $2,3,4,5,6$. The corresponding $k$ values are $12,11,10,9,8$. This yields $5$ cases.

For $i=2$, $j+k=13$. Since $j>2$, $j$ can be $3,4,5,6$. The corresponding $k$ values are $10,9,8,7$. This yields $4$ cases.

For $i=3$, $j+k=12$. Since $j>3$, $j$ can be $4,5$. The corresponding $k$ values are $8,7$. This yields $2$ cases.

For $i=4$, $j+k=11$. Since $j>4$, $j$ can only be $5$. The corresponding $k$ value is $6$. This yields $1$ case.

If we try $i=5$, $j+k=10$. However, since $j$ must be at least $6$, the minimum sum $j+k$ would be $6+7=13$, which is impossible. Thus, we have identified all forbidden cases.

Summing these cases, we get $5+4+2+1=12$. These are the twelve triangles that we must exclude.

Finally, we subtract these from our total:

The beauty of this approach lies in its structure. By breaking down the problem into the total universe and the forbidden zone, we transformed a daunting constraint into a simple, manageable task.

The total number of triangles satisfying the condition is $443$.