1. The problem asks: How many different teams of 3 can be chosen from a squad of 8? 2. This is a combination problem because the order of choosing team members does not matter. 3. The number of ways to choose $k$ items from $n$ items without regard to order is given by the combination formula: $$ \binom{n}{k} = \frac{n!}{k! (n-k)!} $$ 4. Here, $n=8$ and $k=3$. Plugging these into the formula: $$ \binom{8}{3} = \frac{8!}{3! \times (8-3)!} = \frac{8!}{3! \times 5!} $$ 5. Let's evaluate this step by step: $$ 8! = 8 \times 7 \times 6 \times 5! $$ So, $$ \binom{8}{3} = \frac{8 \times 7 \times 6 \times 5!}{3! \times 5!} = \frac{8 \times 7 \times 6}{3!} $$ 6. Calculate $3!$: $$ 3! = 3 \times 2 \times 1 = 6 $$ 7. Therefore: $$ \binom{8}{3} = \frac{8 \times 7 \times 6}{6} $$ 8. Simplify numerator and denominator: $$ \frac{8 \times 7 \times 6}{6} = 8 \times 7 = 56 $$ 9. Final Answer: There are 56 different teams of 3 that can be chosen from 8 people.