and that the number of objects arranged into a 3d tetrahedron of side lengths
$x$
objects are found by the formula
$\sum_{n=1}^x\sum_{m=1}^nm$

how many objects are there when they are arranged into a 4d version of a tetrahedron with side lengths of 5 objects

The answer is 70.

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The number of objects $p$ in a 4d version of a tetrahedron with a side length of $n$ objects is a pentatope number given by the equation $p = {{n + 3} \choose 4} = \frac{n(n + 1)(n + 2)(n + 3)}{24}$ . When $n = 5$ , $p = {8 \choose 4} = \boxed{70}$ .