4th dimensional tetrahedrons (reupload)

Probability Level 5

given that the number of objects arranged into a 2d equilateral triangle of side lengths x x objects are found by the formula ∑ n = 1 x n \sum_{n=1}^xn
and that the number of objects arranged into a 3d tetrahedron of side lengths x x objects are found by the formula ∑ n = 1 x ∑ m = 1 n m \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.

Bento Cheese by bento cheese , 16, USA

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1 solution

David Vreken
Aug 19, 2019

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

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