How many possible combinations of 3 numbers without repetition?

Steps to Solve

1. Identify the combination formula

To find the number of combinations when selecting $r$ items from a set of $n$ items, we use the formula:
$$ C(n, r) = \frac{n!}{r!(n-r)!} $$

Where $n!$ (n factorial) is the product of all positive integers up to $n$.

2. Define the variables

In this problem, we need to decide the values of $n$ (the total number of items in the set) and $r$ (the number of items we want to pick). For this example, let’s assume $n = 10$ (the total numbers available) and $r = 3$ (the three numbers we want to select).

3. Plug the values into the formula

Now we can substitute the values into the combination formula:
$$ C(10, 3) = \frac{10!}{3!(10 - 3)!} $$

4. Calculate factorials

Calculate the factorials involved:
$$ 10! = 10 \times 9 \times 8 \times 7! $$ (we can stop here because $7!$ will cancel out)

$$ 3! = 3 \times 2 \times 1 = 6 $$

$$ 7! = 7! $$ remains as it is since it will cancel out.

5. Simplify the expression

Thus, the expression simplifies to:
$$ C(10, 3) = \frac{10 \times 9 \times 8}{3!} = \frac{10 \times 9 \times 8}{6} $$

6. Perform the calculation

Now we perform the multiplication and division:
First calculate $10 \times 9 \times 8 = 720$.
Then divide that by $6$:
$$ C(10, 3) = \frac{720}{6} = 120 $$