If 8 different flowers can be chosen, how many ways can 4 flowers be selected without regard to order?
Understand the Problem
The question is asking about the number of combinations for selecting 4 flowers from a set of 8 different flowers, where the order of selection does not matter. This can be solved using the combination formula C(n, r) = n! / (r! * (n - r)!) where n is the total number of items to choose from, and r is the number of items to choose.
Answer
$70$
Answer for screen readers
The number of combinations for selecting 4 flowers from a set of 8 different flowers is $70$.
Steps to Solve
- Identify the values of n and r
In this problem, you have 8 different flowers to choose from and you want to select 4 flowers. Therefore, let $n = 8$ and $r = 4$.
- Apply the combination formula
Use the combination formula:
$$ C(n, r) = \frac{n!}{r! \cdot (n - r)!} $$
Substituting our values:
$$ C(8, 4) = \frac{8!}{4! \cdot (8 - 4)!} $$
- Calculate the factorials
Calculate the factorials:
- $8! = 8 \times 7 \times 6 \times 5 \times 4!$
- Since $4!$ appears in both the numerator and the denominator, it can be canceled out:
$$ C(8, 4) = \frac{8 \times 7 \times 6 \times 5}{4!} $$
Now calculate $4! = 4 \times 3 \times 2 \times 1 = 24$.
- Complete the calculation
Now substitute back:
$$ C(8, 4) = \frac{8 \times 7 \times 6 \times 5}{24} $$
Calculate the product in the numerator:
$$ 8 \times 7 \times 6 \times 5 = 1680 $$
Now divide by 24:
$$ C(8, 4) = \frac{1680}{24} = 70 $$
The number of combinations for selecting 4 flowers from a set of 8 different flowers is $70$.
More Information
This problem utilizes the concept of combinations, which are often used in probability and statistics to determine the number of ways to choose items where the order does not matter. It's interesting to note that combinations are widely applied in various fields such as card games, lottery, and combinatorial problems.
Tips
- A common mistake is to confuse combinations with permutations. Remember that in combinations, the order does not matter, while in permutations, it does.
- Another mistake is forgetting to cancel out common terms in factorials, which can lead to unnecessary calculations.
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