To determine how many ways we can choose 2 people out of a group of 10, we can use the combination formula, which is given by:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
where \( n \) is the total number of people, \( r \) is the number of people to choose, and \( ! \) denotes factorial, which is the product of all positive integers up to that number.
In this case, \( n = 10 \) and \( r = 2 \). Thus, we calculate:
\[ \binom{10}{2} = \frac{10!}{2!(10-2)!} = \frac{10!}{2! \cdot 8!} \]
Next, we simplify the factorials:
\[ 10! = 10 \times 9 \times 8! \]
So,
\[ \binom{10}{2} = \frac{10 \times 9 \times 8!}{2! \times 8!} \]
The \( 8! \) in the numerator and denominator cancels out:
\[ \binom{10}{2} = \frac{10 \times 9}{2!} \]
Now we compute \( 2! \):
\[ 2! = 2 \times 1 = 2 \]
Thus, the expression becomes:
\[ \binom{10}{2} = \frac{10 \times 9}{2} = \frac{90}{2} = 45 \]
Therefore, the number of ways to choose 2 people out of a group of 10 is:
\[ \boxed{45} \]