To solve the problem of finding the number of ways to select 3 elements from 10 elements arranged in a circle such that no two are adjacent, we use a case-based approach:

Case 1: The first element is selected

If the first element is chosen, its adjacent elements (2nd and 10th) are excluded. We need to select 2 more non-adjacent elements from positions 3-9 (7 positions, linear arrangement).

The number of ways to choose 2 non-adjacent elements from 7 linear positions is:
[ \binom{7-2+1}{2} = \binom{6}{2} = 15 ]

Case 2: The first element is not selected

We now choose 3 non-adjacent elements from positions 2-10 (9 linear positions).

The number of ways to choose 3 non-adjacent elements from 9 linear positions is:
[ \binom{9-3+1}{3} = \binom{7}{3} = 35 ]

Total Ways

Add the results of both cases:
[ 15 + 35 = 50 ]

Answer: (\boxed{50})

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