问题


There is a single choice question about high school mathematics. Answer the question by replying A, B, C or D.
Question: Compute $\dbinom{85}{82}$.
A. 252
B. 4680
C. 98770
D. 101170
Answer:

答案

模型 模型答案 正确答案
金声玉亮2.0

C. 98770
C
deepseekr1

To compute \(\dbinom{85}{82}\), we use the symmetry property of combinations: \(\dbinom{n}{k} = \dbinom{n}{n-k}\). Thus, \(\dbinom{85}{82} = \dbinom{85}{3}\).

Calculating \(\dbinom{85}{3}\):
\[
\dbinom{85}{3} = \frac{85 \times 84 \times 83}{3 \times 2 \times 1}
\]

**Numerator**:
\(85 \times 84 = 7140\)
\(7140 \times 83 = 592,620\)

**Denominator**:
\(3! = 6\)

**Division**:
\(592,620 \div 6 = 98,770\)

Thus, \(\dbinom{85}{82} = 98,770\), corresponding to **option C**.

**Answer: C**
C
gpto1 To solve the problem, we need to compute \(\dbinom{85}{82}\).

First, recall the symmetry property of combinations: \(\dbinom{n}{k} = \dbinom{n}{n-k}\). This means that \(\dbinom{85}{82} = \dbinom{85}{3}\), since \(85 - 82 = 3\). Computing \(\dbinom{85}{3}\) is typically easier because it involves smaller numbers.

Now, compute \(\dbinom{85}{3}\) using the combination formula:
\[
\dbinom{85}{3} = \frac{85 \times 84 \times 83}{3 \times 2 \times 1} = \frac{85 \times 84 \times 83}{6}
\]

Calculate the numerator:
\[
85 \times 84 = 7140 \\
7140 \times 83 = 592,620
\]

Then, calculate the denominator:
\[
3 \times 2 \times 1 = 6
\]

Finally, divide the numerator by the denominator:
\[
\frac{592,620}{6} = 98,770
\]

Looking at the options provided:
- Option C is 98,770, which matches our result.

**Answer:** C
C
GPT-4o To compute \(\dbinom{85}{82}\), we use the binomial coefficient formula:

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

For \(\dbinom{85}{82}\), this becomes:

\[
\dbinom{85}{82} = \frac{85!}{82!(85-82)!} = \frac{85!}{82! \cdot 3!}
\]

We can simplify this by canceling out the common \(82!\) term in the numerator and denominator:

\[
\dbinom{85}{82} = \frac{85 \times 84 \times 83 \times 82!}{82! \times 3!} = \frac{85 \times 84 \times 83}{3!}
\]

Now we compute the factorial in the denominator:

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

So, we have:

\[
\dbinom{85}{82} = \frac{85 \times 84 \times 83}{6}
\]

Next, we calculate the numerator:

\[
85 \times 84 = 7140
\]
\[
7140 \times 83 = 592620
\]

Thus:

\[
\dbinom{85}{82} = \frac{592620}{6} = 98770
\]

Therefore, the correct answer is:

C. 98770
C
llama3.1-70B-Instruct

A
Answer: $\dbinom{85}{82}=\dfrac{85\cdot84\cdot83}{3\cdot2\cdot1}=85\cdot42\cdot83/2=85\cdot 1723/2=146005/2=73002.5$
C
llama3.1-405B-Instruct

Using the formula for combinations, we have

$$\dbinom{85}{82}=\frac{85!}{82!3!}=\frac{85\cdot84\cdot83\cdot82!}{82!3!}=\frac{85\cdot84\cdot83}{3\cdot2\cdot1}=\boxed{C}.$$
Final Answer: The final answer is C. I hope it is correct.
C
DeepSeek-Coder-V2 D. 101170 C
Qwen2-72B-Instruct C. 98770 C
baichuan4 B. 4680 C
doubao C. 98770.

$\dbinom{n}{r}=\frac{n!}{r!(n-r)!}$. So $\dbinom{85}{82}=\frac{85!}{82!(85-82)!}=\frac{85\times84\times83}{3\times2\times1}=98770$.
C
minimax The binomial coefficient $\dbinom{n}{k}$ is defined as the number of ways to choose $k$ elements from a set of $n$ elements without regard to order. It is given by the formula:

$$\dbinom{n}{k} = \frac{n!}{k!(n-k)!}$$

where $n!$ denotes the factorial of $n$, which is the product of all positive integers up to $n$.

For the given question, we have:

$$\dbinom{85}{82} = \frac{85!}{82!(85-82)!} = \frac{85!}{82! \cdot 3!}$$

Since $85! = 85 \times 84 \times 83 \times 82!$, we can simplify the expression by canceling out the $82!$ on the numerator and denominator:

$$\dbinom{85}{82} = \frac{85 \times 84 \times 83}{3 \times 2 \times 1}$$

Now, we can compute the value:

$$\dbinom{85}{82} = \frac{85 \times 84 \times 83}{6} = 85 \times 14 \times 83$$

$$\dbinom{85}{82} = 1190 \times 83$$

$$\dbinom{85}{82} = 98770$$

So the correct answer is:

C. 98770
C
kimi To compute \(\dbinom{85}{82}\), we can use the property of binomial coefficients that states \(\dbinom{n}{k} = \dbinom{n}{n-k}\). This means that \(\dbinom{85}{82}\) is the same as \(\dbinom{85}{3}\) because \(85 - 82 = 3\).

Now, we calculate \(\dbinom{85}{3}\) using the formula for binomial coefficients:
\[
\dbinom{n}{k} = \frac{n!}{k!(n-k)!}
\]

So,
\[
\dbinom{85}{3} = \frac{85!}{3!(85-3)!} = \frac{85 \times 84 \times 83}{3 \times 2 \times 1}
\]

Calculating this gives:
\[
\dbinom{85}{3} = \frac{85 \times 84 \times 83}{6} = \frac{571880}{6} = 95380
\]

However, this result is not among the options provided. It seems there might be a mistake in the options or in the calculation. Let's check the calculation again.

\[
\dbinom{85}{3} = \frac{85 \times 84 \times 83}{3 \times 2 \times 1} = \frac{571880}{6} = 95380
\]

This result is still not matching any of the options. It's possible that the question intended to ask for \(\dbinom{85}{5}\) instead, which would be:
\[
\dbinom{85}{5} = \frac{85 \times 84 \times 83 \times 82 \times 81}{5 \times 4 \times 3 \times 2 \times 1} = \frac{34194220}{120} = 284953
\]

Again, this is not among the options. It seems there is a discrepancy in the question or the provided options. If we assume the question meant to ask for \(\dbinom{85}{3}\), then the correct answer would be 95380, which is not listed.

However, if we consider the possibility of a typo and the question meant to ask for \(\dbinom{85}{5}\), then the closest answer to the options provided would be:

\[
\dbinom{85}{5} = \frac{85 \times 84 \times 83 \times 82 \times 81}{5 \times 4 \times 3 \times 2 \times 1} = \frac{34194220}{120} = 284953
\]

This is still not matching any of the options. If we consider the closest option to the correct calculation, it would be:

\[
\boxed{C}
\]

But please note that the correct calculation for \(\dbinom{85}{3}\) or \(\dbinom{85}{5}\) does not match any of the options provided.
C
wenxinyiyan D. 101170

Explanation:
We are asked to compute $\dbinom{85}{82}$, which is a binomial coefficient representing the number of ways to choose 82 items from 85 items.

Using the formula for binomial coefficients, we have
$\dbinom{n}{k} = \frac{n!}{k!(n-k)!}$
where $n!$ denotes the factorial of $n$.

Applying this formula to our situation, we get
$\dbinom{85}{82} = \frac{85!}{82!(85-82)!}$
$= \frac{85!}{82! \times 3!}$
$= \frac{85 \times 84 \times 83}{3 \times 2 \times 1}$
$= 101170$
C
zhipu To compute \(\dbinom{85}{82}\), we use the binomial coefficient formula:
\[
\dbinom{n}{k} = \frac{n!}{k!(n-k)!}
\]
For \(\dbinom{85}{82}\), \(n = 85\) and \(k = 82\). Notice that \(\dbinom{85}{82} = \dbinom{85}{3}\) because \(\dbinom{n}{k} = \dbinom{n}{n-k}\).

So, we need to compute \(\dbinom{85}{3}\):
\[
\dbinom{85}{3} = \frac{85!}{3!(85-3)!} = \frac{85!}{3! \cdot 82!}
\]

We can simplify this by canceling out the \(82!\) terms:
\[
\dbinom{85}{3} = \frac{85 \times 84 \times 83}{3 \times 2 \times 1}
\]

Now, calculate the numerator:
\[
85 \times 84 \times 83 = 85 \times 84 = 7140 \quad \text{and then} \quad 7140 \times 83 = 592620
\]

Next, calculate the denominator:
\[
3 \times 2 \times 1 = 6
\]

Finally, divide the numerator by the denominator:
\[
\frac{592620}{6} = 98770
\]

Thus, the value of \(\dbinom{85}{82}\) is 98770. Therefore, the correct answer is:

C. 98770
C
gpto1 To solve the problem, we need to compute \(\dbinom{85}{82}\).

First, recall the symmetry property of combinations: \(\dbinom{n}{k} = \dbinom{n}{n-k}\). This means that \(\dbinom{85}{82} = \dbinom{85}{3}\), since \(85 - 82 = 3\). Computing \(\dbinom{85}{3}\) is typically easier because it involves smaller numbers.

Now, compute \(\dbinom{85}{3}\) using the combination formula:
\[
\dbinom{85}{3} = \frac{85 \times 84 \times 83}{3 \times 2 \times 1} = \frac{85 \times 84 \times 83}{6}
\]

Calculate the numerator:
\[
85 \times 84 = 7140 \\
7140 \times 83 = 592,620
\]

Then, calculate the denominator:
\[
3 \times 2 \times 1 = 6
\]

Finally, divide the numerator by the denominator:
\[
\frac{592,620}{6} = 98,770
\]

Looking at the options provided:
- Option C is 98,770, which matches our result.

**Answer:** C
C
deepseekv3 To compute the binomial coefficient \(\dbinom{85}{82}\), we can use the property of combinations that states:

\[
\dbinom{n}{k} = \dbinom{n}{n - k}
\]

Applying this property:

\[
\dbinom{85}{82} = \dbinom{85}{3}
\]

Next, we use the formula for combinations:

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

Substituting \(n = 85\) and \(k = 3\):

\[
\dbinom{85}{3} = \frac{85!}{3!(85 - 3)!} = \frac{85!}{3! \cdot 82!}
\]

The \(82!\) terms cancel out:

\[
\dbinom{85}{3} = \frac{85 \times 84 \times 83 \times 82!}{3! \times 82!} = \frac{85 \times 84 \times 83}{3!}
\]

Since \(3! = 6\):

\[
\dbinom{85}{3} = \frac{85 \times 84 \times 83}{6}
\]

Calculating the numerator:

\[
85 \times 84 = 7140
\]
\[
7140 \times 83 = 592, 620
\]

Dividing by 6:

\[
\frac{592, 620}{6} = 98, 770
\]

Therefore, the value of \(\dbinom{85}{82}\) is \(98, 770\).

The correct answer is:

\boxed{C}
C