How many possible combinations of 3 numbers without repetition?
Understand the Problem
The question is asking for the number of combinations possible when selecting 3 numbers from a set, without allowing the same number to appear more than once. This will require an understanding of combination formulas typically used in combinatorics.
Answer
The number of combinations is $120$.
Answer for screen readers
The number of combinations possible when selecting 3 numbers from a set of 10 is $120$.
Steps to Solve
- 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 $n$.
- 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, letâ€™s assume $n = 10$ (the total numbers available) and $r = 3$ (the three numbers we want to select).
- Plug the values into the formula
Now we can substitute the values into the combination formula:
$$ C(10, 3) = \frac{10!}{3!(10 - 3)!} $$
- 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.
- Simplify the expression
Thus, the expression simplifies to:
$$ C(10, 3) = \frac{10 \times 9 \times 8}{3!} = \frac{10 \times 9 \times 8}{6} $$
- Perform the calculation
Now we perform the multiplication and division:
First calculate $10 \times 9 \times 8 = 720$.
Then divide that by $6$:
$$ C(10, 3) = \frac{720}{6} = 120 $$
The number of combinations possible when selecting 3 numbers from a set of 10 is $120$.
More Information
In combinatorics, combinations refer to the selection of items where the order does not matter. This formula helps in determining how many different groups can be formed from a larger set, essential in probability and statistics.
Tips
- Forgetting to use the combination formula instead of the permutation formula.
- Miscalculating factorials, especially the fact that $0! = 1$.
- Not recognizing that order does not matter in combinations.