<Question>
Compute $\dbinom{31}{28}$.
</Question>

<think>
<Plan_1>
Okay, so I need to compute the combination \(\dbinom{31}{28}\). Hmm, combinations... right, the formula is \(\dbinom{n}{k} = \frac{n! }{k! (n - k)! }\). Let me make sure I remember that correctly. Factorials can get really big, so maybe there's a way to simplify this before calculating everything.

<prompt_1.1> Set up the combination formula for \(\dbinom{31}{28}\) using the formula \(\dbinom{n}{k} = \frac{n! }{k! (n - k)! }\). </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
First, let me recall that \(\dbinom{n}{k}\) represents the number of ways to choose \(k\) elements from a set of \(n\) without considering the order. But in this case, the numbers are 31 and 28. Wait, 28 is larger than 31 minus 28, which is 3. Oh, right! Sometimes it's easier to compute the smaller of \(k\) and \(n - k\) because the combination has the same value as \(\dbinom{n}{n - k}\). Let me check that. Since 31 - 28 = 3, which is smaller than 28, maybe I can compute \(\dbinom{31}{3}\) instead. That should be the same.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
So, \(\dbinom{31}{28} = \dbinom{31}{3}\). That might be easier because 3 is a smaller number, and the factorial calculations will be more manageable.

Let me confirm that property of combinations.
Yes, \(\dbinom{n}{k} = \dbinom{n}{n - k}\) because choosing \(k\) elements to include is the same as choosing \(n - k\) elements to exclude. So, this symmetry can simplify calculations.

Alright, so now I can compute \(\dbinom{31}{3}\) instead. Let's write out the formula for that.

<prompt_2.1> Calculate \(\dbinom{31}{3}\) using the simplified formula \(\frac{31 \times 30 \times 29}{3 \times 2 \times 1}\). </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
\[
\dbinom{31}{3} = \frac{31! }{3! \cdot 28! }
\]

But since \(31! = 31 \times 30 \times 29 \times 28! \), the \(28! \) in the numerator and denominator will cancel out. So that simplifies to:

\[
\frac{31 \times 30 \times 29}{3 \times 2 \times 1}
\]

Because \(31! / 28! \) leaves 31 × 30 × 29, and then we divide by 3!
which is 6.
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Let me compute the numerator first.

<prompt_3.1> Calculate the numerator and denominator of the simplified expression and then divide to find the final result. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
31 multiplied by 30 is 930. Then, 930 multiplied by 29. Hmm, 930 × 29. Let me break that down. 930 × 30 would be 27,900, right? But since it's 930 × 29, that's 30 less than 30. So 27,900 minus 930 is 26,970. So the numerator is 26,970.

Now the denominator is 3 × 2 × 1, which is 6. So 26,970 divided by 6.
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
Let me do that division. 26,970 ÷ 6. Well, 6 goes into 26 four times (6 × 4 = 24), remainder 2. Then bring down the 9 to make 29.

<prompt_4.1> Divide 26,970 by 6 to find the final result. </prompt_4.1>
-----

Alternatively, let's check with another approach. Let me compute the original formula without simplifying first.

<prompt_4.2> Recompute \(\dbinom{31}{28}\) using the original formula \(\frac{31! }{28! \cdot 3! }\) to verify the result. </prompt_4.2>
</Plan_4>

<Execution_4>
<execution_4.1>
6 into 29 is 4 times again (6 × 4 = 24), remainder 5. Bring down the 7 to make 57. 6 into 57 is 9 times (6 × 9 = 54), remainder 3. Bring down the 0. 6 into 30 is 5 times exactly. So putting it all together: 4, 4, 9, 5. Wait, let me check that again. The first division step: 26 ÷ 6 is 4, remainder 2. Then 29 ÷ 6 is 4, remainder 5. Then 57 ÷ 6 is 9, remainder 3. Then 30 ÷ 6 is 5. So the result is 4,495.
</execution_4.1>

<execution_4.2>
\[
\dbinom{31}{28} = \frac{31! }{28! \cdot 3! }
\]

As mentioned, 31! / 28! = 31 × 30 × 29. Then, dividing by 3! which is 6. So, 31 × 30 × 29 / 6. Let's compute 30 / 6 first, which is 5. Then, 31 × 5 × 29. So 31 × 5 is 155, then 155 × 29. Let's compute 155 × 30 first, which is 4,650, then subtract 155 to get 155 × 29. 4,650 - 155 = 4,495. Wait, that's different from before. Wait, what? Wait, no, that can't be. Wait, 31 × 5 × 29. Let me do it step by step.

First, 31 × 5 = 155. Then, 155 × 29. Let's break down 29 into 30 - 1. So, 155 × 30 = 4,650, and 155 × 1 = 155. So, 4,650 - 155 = 4,495. Wait, but previously, when I did 26,970 ÷ 6, I got 4,495. So that's the same. So both methods give 4,495.
</execution_4.2>
</Execution_4>

<Plan_5>
Based on execution_4.1:
Wait, but let me check that division again to be sure.

<prompt_5.1> Verify the division of 26,970 by 6 to confirm the result of the combination. </prompt_5.1>
-----

Based on execution_4.2:
Alternatively, if I compute 31 × 30 × 29 first, which is 31 × 30 = 930, then 930 × 29. Let me compute 930 × 29 again.

<prompt_5.2> Recheck the calculations using different methods to ensure the accuracy of the result. </prompt_5.2>

Let me just verify once more with a different approach. Let's use the multiplicative formula for combinations.

<prompt_5.3> Verify the result using the multiplicative formula for combinations and confirm the final answer. </prompt_5.3>
</Plan_5>

<Execution_5>
<execution_5.1>
26,970 divided by 6: 6 × 4,000 is 24,000. Subtract that from 26,970: 2,970. Then 6 × 495 is 2,970. So 4,000 + 495 = 4,495.
</execution_5.1>

<execution_5.2>
29 can be written as 30 - 1, so 930 × 30 = 27,900, then subtract 930: 27,900 - 930 = 26,970. Then divide by 6: 26,970 ÷ 6. Let's divide 26,970 by 6 step by step. 6 into 2 is 0, 6 into 26 is 4 (4×6=24), remainder 2. Bring down 9: 29. 6 into 29 is 4 (4×6=24), remainder 5. Bring down 7: 57. 6 into 57 is 9 (9×6=54), remainder 3. Bring down 0: 30. 6 into 30 is 5. So the result is 4,495. So that seems correct.
</execution_5.2>

<execution_5.3>
The formula is:

\[
\dbinom{n}{k} = \frac{n \times (n - 1) \times \dots \times (n - k + 1)}{k! }
\]

So here, n = 31, k = 28. So, the numerator is 31 × 30 × 29 × ... × (31 - 28 + 1) = 31 × 30 × 29 × ... × 4. Wait, but that's a huge product. However, since we are dividing by 28! , most of the terms will cancel out. Wait, but in this case, since k = 28, the formula would be:

\[
\dbinom{31}{28} = \frac{31 \times 30 \times 29 \times 28! }{28! \times 3! }
\]

Ah, right! So the 28! cancels out, leaving us with \(\frac{31 \times 30 \times 29}{3! }\), which is the same as before. So, this confirms again that it's \(\frac{31 \times 30 \times 29}{6}\), leading to 4,495.

I think all approaches lead to the same answer. Let me just check with another angle. Suppose I use the recursive formula for combinations: \(\dbinom{n}{k} = \dbinom{n - 1}{k - 1} + \dbinom{n - 1}{k}\). But applying that here might not be helpful because it would require computing many smaller combinations, which could be tedious. However, maybe for verification, let's see:

\(\dbinom{31}{28} = \dbinom{30}{27} + \dbinom{30}{28}\). Then, \(\dbinom{30}{27} = \dbinom{30}{3}\), which is \(\frac{30 \times 29 \times 28}{6} = \frac{30 \times 29 \times 28}{6}\). Let's compute that: 30 ÷ 6 = 5, so 5 × 29 × 28. 5 × 29 = 145, then 145 × 28. 145 × 20 = 2,900; 145 × 8 = 1,160. Sum is 2,900 + 1,160 = 4,060.

Then, \(\dbinom{30}{28} = \dbinom{30}{2}\). Which is \(\frac{30 × 29}{2} = 15 × 29 = 435\). So adding those two: 4,060 + 435 = 4,495. That matches the previous result. So, that's another way to confirm it's 4,495.

Another check: Let's compute \(\dbinom{31}{3}\) directly. 31 choose 3. Using the formula:

\[
\dbinom{31}{3} = \frac{31 × 30 × 29}{3 × 2 × 1} = \frac{31 × 30 × 29}{6}
\]

As before, 30 ÷ 6 = 5, so 31 × 5 × 29. 31 × 5 = 155, 155 × 29. Let's compute 155 × 30 = 4,650, subtract 155 to get 4,650 - 155 = 4,495. Yep, same answer. So all these different methods converge to 4,495.

Therefore, I can be confident that the value of \(\dbinom{31}{28}\) is 4,495.
</execution_5.3>
</Execution_5>

<Final_answer>
Therefore, $\dbinom{31}{28} = \dbinom{31}{3} = \boxed{4495}$.
</Final_answer>
</think>