问题

You are supposed to provide a solution to a given problem.

Problem:
In how many ways can 3 students be selected from a class of 20 to form a study group?
Solution:

答案

模型 模型答案 正确答案
金声玉亮2.0 To solve this problem, we need to use the concept of combinations. A combination is a selection of items without regard to the order in which they are selected. In this case, we want to select 3 students out of 20 without regard to the order in which they are selected.

The formula for combinations is:

C(n, k) = n! / (k!(n-k)!)

Where:
- n is the total number of items (in this case, students) to choose from,
- k is the number of items (students) to choose,
- "!" denotes factorial, which means the product of all positive integers up to that number (e.g., 5! = 5 x 4 x 3 x 2 x 1 = 120).

Using this formula, we can calculate the number of ways to select 3 students out of 20 as follows:

C(20, 3) = 20! / (3!(20-3)!)
= 20! / (3!17!)
= (20 x 19 x 18) / (3 x 2 x 1)
= 1140

Therefore, there are 1140 ways to select 3 students out of a class of 20 to form a study group.		 1140
deepseekr1

To determine the number of ways to select 3 students from a class of 20, we use combinations since the order of selection does not matter. The formula for combinations is:

\[
C(n, k) = \frac{n!}{k! \cdot (n - k)!}
\]

Plugging in \(n = 20\) and \(k = 3\):

\[
C(20, 3) = \frac{20!}{3! \cdot 17!} = \frac{20 \times 19 \times 18 \times \cancel{17!}}{3! \times \cancel{17!}} = \frac{20 \times 19 \times 18}{3!}
\]

Simplifying the denominator (\(3! = 6\)) and numerator:

\[
\frac{20 \times 19 \times 18}{6} = \frac{6840}{6} = 1140
\]

**Answer:** There are \(\boxed{1140}\) ways to select 3 students from a class of 20.		 1140
gpto1 To find the number of ways to select 3 students from a class of 20 to form a study group, we need to calculate the number of combinations of 20 students taken 3 at a time. This is a classic combinatorial problem.

The formula for combinations (also known as "n choose k") is:

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

Where:
- \( n \) is the total number of items (students, in this case),
- \( k \) is the number of items to choose,
- \( n! \) is the factorial of \( n \).

Plugging in the values:

\[
C(20, 3) = \frac{20!}{3!(20 - 3)!}
\]

Simplify the factorial expression:

\[
C(20, 3) = \frac{20 \times 19 \times 18 \times 17!}{(3 \times 2 \times 1) \times 17!}
\]

Notice that \( 17! \) cancels out from the numerator and the denominator:

\[
C(20, 3) = \frac{20 \times 19 \times 18}{3 \times 2 \times 1}
\]

Compute the numerator and the denominator:

\[
C(20, 3) = \frac{6840}{6}
\]

\[
C(20, 3) = 1140
\]

**Answer:**

There are **1140 ways** to select 3 students from a class of 20 to form a study group.		 1140
GPT-4o To determine the number of ways to select 3 students from a class of 20 to form a study group, we use the concept of combinations. The formula for combinations is given by:

\[ C(n, r) = \frac{n!}{r!(n-r)!} \]

where \( n \) is the total number of items (or students, in this case), \( r \) is the number of items to choose, and \( ! \) denotes factorial, which is the product of all positive integers up to that number.

For this problem:
- \( n = 20 \) (total number of students)
- \( r = 3 \) (number of students to be selected)

Plugging these values into the formula, we get:

\[ C(20, 3) = \frac{20!}{3!(20-3)!} = \frac{20!}{3! \cdot 17!} \]

Now, we simplify this expression. The factorial \( 20! \) (20 factorial) is the product of all positive integers up to 20, and similarly for \( 17! \) and \( 3! \):

\[ 20! = 20 \times 19 \times 18 \times 17! \]

So,

\[ \frac{20!}{3! \cdot 17!} = \frac{20 \times 19 \times 18 \times 17!}{3! \times 17!} \]

Since \( 17! \) appears in both the numerator and the denominator, they cancel each other out:

\[ \frac{20 \times 19 \times 18}{3!} \]

Next, we compute \( 3! \) (3 factorial):

\[ 3! = 3 \times 2 \times 1 = 6 \]

Thus, the expression simplifies to:

\[ \frac{20 \times 19 \times 18}{6} \]

Now we perform the multiplication and division:

\[ 20 \times 19 = 380 \]
\[ 380 \times 18 = 6840 \]
\[ \frac{6840}{6} = 1140 \]

Therefore, the number of ways to select 3 students from a class of 20 is:

\[ 1140 \]

So, there are \( \boxed{1140} \) ways to select 3 students from a class of 20 to form a study group.		 1140
llama3.1-70B-Instruct

To solve this problem, we can use the concept of combinations. Since the order of selection does not matter (i.e., selecting student A, B, and C is the same as selecting student C, B, and A), we can use the combination formula:

C(n, k) = n! / (k!(n-k)!)

where n is the total number of students (20) and k is the number of students to be selected (3).

Plugging in the values, we get:

C(20, 3) = 20! / (3!(20-3)!)
= 20! / (3!17!)
= (20 × 19 × 18) / (3 × 2 × 1)
= 1140

Therefore, there are 1140 ways to select 3 students from a class of 20 to form a study group.		 1140
llama3.1-405B-Instruct

To select 3 students from a class of 20, we can use the combination formula:

C(n, k) = n! / (k!(n-k)!)

where n is the total number of students (20), k is the number of students to be selected (3), and ! denotes the factorial function.

Plugging in the values, we get:

C(20, 3) = 20! / (3!(20-3)!)
= 20! / (3!17!)
= (20 × 19 × 18) / (3 × 2 × 1)
= 1140

Therefore, there are 1140 ways to select 3 students from a class of 20 to form a study group.		 1140
DeepSeek-Coder-V2 To solve the problem of selecting 3 students from a class of 20 to form a study group, we need to use the concept of combinations. Combinations are used when the order of selection does not matter.

The formula for combinations is given by:
\[
C(n, k) = \frac{n!}{k!(n-k)!}
\]
where \( n \) is the total number of items to choose from, \( k \) is the number of items to choose, and \( ! \) denotes factorial, which is the product of all positive integers up to that number.

In this problem, \( n = 20 \) and \( k = 3 \). Plugging these values into the formula, we get:
\[
C(20, 3) = \frac{20!}{3!(20-3)!} = \frac{20!}{3! \cdot 17!}
\]

Next, we simplify the factorials. Notice that \( 20! \) can be written as \( 20 \times 19 \times 18 \times 17! \). Since \( 17! \) appears in both the numerator and the denominator, they cancel each other out:
\[
C(20, 3) = \frac{20 \times 19 \times 18 \times 17!}{3! \times 17!} = \frac{20 \times 19 \times 18}{3!}
\]

Now, we calculate \( 3! \):
\[
3! = 3 \times 2 \times 1 = 6
\]

So, the expression simplifies to:
\[
C(20, 3) = \frac{20 \times 19 \times 18}{6}
\]

We can now perform the multiplication and division:
\[
20 \times 19 = 380
\]
\[
380 \times 18 = 6840
\]
\[
\frac{6840}{6} = 1140
\]

Therefore, the number of ways to select 3 students from a class of 20 is:
\[
\boxed{1140}
\]		 1140
Qwen2-72B-Instruct To solve this problem, we need to use the concept of combinations. A combination is a selection of items without regard to the order in which they are selected. In this case, we want to select 3 students out of 20 without regard to the order in which they are selected.

The formula for combinations is:

C(n, k) = n! / (k!(n-k)!)

Where:
- n is the total number of items (in this case, students) to choose from,
- k is the number of items (students) to choose,
- "!" denotes factorial, which means the product of all positive integers up to that number (e.g., 5! = 5 x 4 x 3 x 2 x 1 = 120).

Using this formula, we can calculate the number of ways to select 3 students out of 20 as follows:

C(20, 3) = 20! / (3!(20-3)!)
= 20! / (3!17!)
= (20 x 19 x 18) / (3 x 2 x 1)
= 1140

Therefore, there are 1140 ways to select 3 students out of a class of 20 to form a study group.		 1140
baichuan4 To find the number of ways 3 students can be selected from a class of 20, we can use the formula for combinations:

C(n, k) = n! / (k! * (n - k)!)

where:
- n is the total number of items (in this case, students)
- k is the number of items to select
- "!" denotes the factorial operation (e.g., 5! = 5 * 4 * 3 * 2 * 1)

In this problem, n = 20 (total students) and k = 3 (students to select for the study group).

Let's calculate:

C(20, 3) = 20! / (3! * (20 - 3)!)
= 20! / (3! * 17!)
= (20 * 19 * 18 * 17!) / (3 * 2 * 1 * 17!)
= (20 * 19 * 18) / (3 * 2 * 1)
= 1140

So, there are 1140 different ways to select 3 students from a class of 20 to form a study group.		 1140
doubao The number of ways to select 3 students from a class of 20 can be found using the combination formula: C(20,3) = 20! / (3! * (20-3)!) = 20! / (3! * 17!) = (20 * 19 * 18) / (3 * 2 * 1) = 1140. So, there are 1140 ways. 1140
minimax To solve this problem, we can use the combination formula, which is used to determine the number of ways to choose k items from a set of n items without regard to the order in which they are selected. The formula for combinations is given by:

C(n, k) = n! / (k! * (n - k)!)

Where:
- C(n, k) is the number of combinations,
- n is the total number of items,
- k is the number of items to choose,
- n! denotes the factorial of n, which is the product of all positive integers from 1 to n.

In this case, we want to select 3 students (k = 3) from a class of 20 (n = 20). Plugging these values into the combination formula gives us:

C(20, 3) = 20! / (3! * (20 - 3)!)
= 20! / (3! * 17!)
= (20 * 19 * 18 * 17!) / (3 * 2 * 1 * 17!)
= (20 * 19 * 18) / (3 * 2 * 1)
= (20 * 19 * 3) / (3 * 2)
= (20 * 19) / 2
= 380 / 2
= 190

Therefore, there are 190 different ways to select 3 students from a class of 20 to form a study group.		 1140
kimi To solve this problem, we can use the concept of combinations in mathematics. Combinations are used when the order of selection does not matter.

The formula for combinations is:

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

where:
- \( n \) is the total number of items to choose from (in this case, 20 students),
- \( k \) is the number of items to choose (in this case, 3 students),
- \( n! \) denotes the factorial of \( n \) (the product of all positive integers from 1 to \( n \)),
- \( k! \) denotes the factorial of \( k \).

For this problem, we want to find \( C(20, 3) \):

\[ C(20, 3) = \frac{20!}{3!(20-3)!} \]
\[ C(20, 3) = \frac{20!}{3! \cdot 17!} \]
\[ C(20, 3) = \frac{20 \times 19 \times 18 \times 17!}{3 \times 2 \times 1 \times 17!} \]
\[ C(20, 3) = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} \]
\[ C(20, 3) = \frac{6840}{6} \]
\[ C(20, 3) = 1140 \]

So, there are 1140 different ways to select 3 students from a class of 20 to form a study group.		 1140
wenxinyiyan 这个问题是一个组合问题，即从20个学生中选择3个学生，不考虑顺序。在数学中，从n个不同元素中，任取m(m≤n)个元素并成一组，叫做从n个不同元素中取出m个元素的一个组合。组合的数学公式表示为 C(n, m)，也可以写作 nCm，其中 n 是总的元素数量，m 是要选择的元素数量。

在这个问题中，n=20（总的学生数量）, m=3（要选择的学生数量）。所以我们需要计算 C(20, 3)。

组合公式 C(n, m) = n! / (m!(n-m)!)，其中 "!" 表示阶乘，即一个数与所有小于它的正整数的乘积。

将n=20, m=3代入公式，我们可以得到 C(20, 3) = 20! / (3!17!) = 1140。

所以，从20个学生中选择3个学生形成学习小组的方式有1140种。		 1140
zhipu To solve this problem, we use the concept of combinations. A combination is a selection of items where the order does not matter. The number of ways to choose \( k \) items from a set of \( n \) items is given by the combination formula:

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

Where:
- \( n! \) (n factorial) is the product of all positive integers up to \( n \),
- \( k \) is the number of items to choose,
- \( n-k \) is the number of items that are not chosen.

In this problem, we have \( n = 20 \) students and we want to choose \( k = 3 \) students to form a study group. Plugging these values into the combination formula gives us:

\[
^{20}C_3 = \frac{20!}{3!(20-3)!} = \frac{20!}{3! \cdot 17!}
\]

Now let's simplify this expression. We know that \( 20! \) (20 factorial) is \( 20 \times 19 \times 18 \times 17! \), and since we have \( 17! \) in both the numerator and the denominator, they cancel each other out:

\[
^{20}C_3 = \frac{20 \times 19 \times 18}{3!}
\]

We also know that \( 3! = 3 \times 2 \times 1 = 6 \), so we can substitute that in:

\[
^{20}C_3 = \frac{20 \times 19 \times 18}{6}
\]

Now we just perform the arithmetic:

\[
^{20}C_3 = \frac{20 \times 19 \times 18}{6} = \frac{6840}{6} = 1140
\]

So, there are 1140 ways to select 3 students from a class of 20 to form a study group.		 1140
gpto1 To find the number of ways to select 3 students from a class of 20 to form a study group, we need to calculate the number of combinations of 20 students taken 3 at a time. This is a classic combinatorial problem.

The formula for combinations (also known as "n choose k") is:

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

Where:
- \( n \) is the total number of items (students, in this case),
- \( k \) is the number of items to choose,
- \( n! \) is the factorial of \( n \).

Plugging in the values:

\[
C(20, 3) = \frac{20!}{3!(20 - 3)!}
\]

Simplify the factorial expression:

\[
C(20, 3) = \frac{20 \times 19 \times 18 \times 17!}{(3 \times 2 \times 1) \times 17!}
\]

Notice that \( 17! \) cancels out from the numerator and the denominator:

\[
C(20, 3) = \frac{20 \times 19 \times 18}{3 \times 2 \times 1}
\]

Compute the numerator and the denominator:

\[
C(20, 3) = \frac{6840}{6}
\]

\[
C(20, 3) = 1140
\]

**Answer:**

There are **1140 ways** to select 3 students from a class of 20 to form a study group.		 1140
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