8 choose 8
Understand the Problem
The question is asking for the mathematical calculation of '8 choose 8', which is a combinatorial function represented as C(n, k) or nCk. This function calculates the number of ways to choose k elements from a set of n elements, and in this case, selecting all elements from a set of 8.
Answer
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Answer for screen readers
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Steps to Solve
- Identify the formula for combinations
The formula for combinations, or "n choose k", is given by:
$$ C(n, k) = \frac{n!}{k! \cdot (n-k)!} $$
where $n!$ (n factorial) is the product of all positive integers up to $n$.
- Substitute the values of n and k
In this case, we have $n = 8$ and $k = 8$. Let's substitute these values into the formula:
$$ C(8, 8) = \frac{8!}{8! \cdot (8-8)!} $$
- Simplify the factorials
We know $8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$. However, $(8-8)!$ simplifies to $0!$, and by definition, $0! = 1$.
So the expression becomes:
$$ C(8, 8) = \frac{8!}{8! \cdot 1} $$
- Calculate the value of the expression
Since the numerator and denominator are identical ($8!$), we can simplify:
$$ C(8, 8) = \frac{8!}{8!} = 1 $$
This means that there is exactly one way to choose all 8 elements from a set of 8.
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More Information
The result of '8 choose 8' is 1, which signifies that there is exactly one way to select all items from a set. This concept is foundational in combinatorics, illustrating that choosing all items from a set can only be done in a single way.
Tips
- Confusing the selection process: Sometimes, people misinterpret combinations to mean arrangements, but here we just need the number of ways to choose, not the order.
- Ignoring the value of $0!$: Remember that $0!$ is always 1, which is crucial in calculations involving combinations.
