Exploring Pascal’s Pyramid

A triangular pyramid is constructed using spherical balls so that each ball rests on exactly three balls of the next lower level.

Then, we calculate the number of paths leading from the apex to each position:
A path starts at the apex and progresses downwards to any of the three spheres directly below the current position.
Consequently, the number of paths to reach a certain position is the sum of the numbers immediately above it (depending on the position, there are up to three numbers above it).
The result is Pascal’s pyramid and the numbers at each level $n$ are the coefficients of the trinomial expansion
$(x + y + z)^n$.
How many coefficients in the expansion of $(x + y + z)^{200000}$ are multiples of $10^{12}$?

This is a quite advanced problem involving combinatorics, binomial coefficients, and number theory. It seems to involve sums of binomial coefficients in a trinomial expansion, which are given by trinomial coefficients (or, in this case, coefficients of the tetrahedron or Pascal’s pyramid), and a requirement that these be multiples of a power of 10 (in this case, $10^{12}$).

However, there is no known direct formula or a straightforward way of computing the number of trinomial coefficients in $(x+y+z)^n$ that are multiples of $10^{12}$. This is beyond the standard curriculum and requires a deep investigation in combinatorics and number theory.

The solution to this problem may require generating functions or considerations of properties of binomial coefficients modulo $10^{12}$ or modulo prime factors of $10^{12}$, and could require advanced number-theoretic techniques or possibly even computer calculations. Such a computation will be difficult and time-consuming.

If this is a problem posed in an academic context, it is suggested that you consult your professors or peers who are knowledgeable in these areas. Some specific results about trinomial coefficients, and about binomial coefficients modulo powers of primes, are known and could potentially be useful, but a complete solution to your problem is likely to be quite involved.

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