To determine how many different selections of four people can be chosen from eight, we can use the binomial coefficient, commonly denoted as \( \binom{n}{r} \), which represents the number of ways to choose \( r \) elements from a set of \( n \) elements without regard to the order of selection. In this case, we need to calculate \( \binom{8}{4} \).
First, let's construct Pascal's Triangle up to row 8:
Row 0: 1
Row 1: 1 1
Row 2: 1 2 1
Row 3: 1 3 3 1
Row 4: 1 4 6 4 1
Row 5: 1 5 10 10 5 1
Row 6: 1 6 15 20 15 6 1
Row 7: 1 7 21 35 35 21 7 1
Row 8: 1 8 28 56 70 28 8 1
In Pascal's Triangle, each entry \( \binom{n}{r} \) corresponds to the \( r \)-th entry of the \( n \)-th row, starting from row 0.
- Row 8:
- \( \binom{8}{0} = 1 \)
- \( \binom{8}{1} = 8 \)
- \( \binom{8}{2} = 28 \)
- \( \binom{8}{3} = 56 \)
- \( \binom{8}{4} = 70 \)
- \( \binom{8}{5} = 56 \)
- \( \binom{8}{6} = 28 \)
- \( \binom{8}{7} = 8 \)
- \( \binom{8}{8} = 1 \)
From the constructed Pascal's Triangle, we find that \( \binom{8}{4} = 70 \).
Thus, the total number of different selections of four out of eight people that Cherese can choose is \( \boxed{70} \).