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🗣 Wd提问: ip group

🤖 Ai回答:
To solve the problem, we need to determine the number of ways to form a group of 5 people from a set of 10 people、This is a classic combination problem, where the order of selection does not matter、The number of ways to choose \( k \) elements from a set of \( n \) elements is given by the combination formula:

\[
\binom{n}{k} = \frac{n!}{k!(n-k)!}
\]

In this problem, \( n = 10 \) and \( k = 5 \)、Plugging these values into the formula, we get:

\[
\binom{10}{5} = \frac{10!}{5!5!}
\]

Next, we need to compute the factorials、Recall that \( n! \) (n factorial) is the product of all positive integers up to \( n \)、So, we have:

\[
10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1
\]
\[
5! = 5 \times 4 \times 3 \times 2 \times 1
\]

However, we don't need to fully expand \( 10! \) because many terms will cancel out、Instead, we can write:

\[
10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5!
\]

So the expression for the combination becomes:

\[
\binom{10}{5} = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5!}{5! \times 5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1}
\]

Now, we can cancel out the \( 5! \) in the numerator and the denominator:

\[
\binom{10}{5} = \frac{10 \times 9 \times 8 \times 7 \times 6}{120}
\]

Next, we perform the multiplication in the numerator:

\[
10 \times 9 = 90
\]
\[
90 \times 8 = 720
\]
\[
720 \times 7 = 5040
\]
\[
5040 \times 6 = 30240
\]

So the expression is now:

\[
\binom{10}{5} = \frac{30240}{120}
\]

Finally, we perform the division:

\[
\frac{30240}{120} = 252
\]

Therefore, the number of ways to form a group of 5 people from a set of 10 people is:

\[
\boxed{252}
\]

Ai作答

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