Consider the expansion of `( 1+ x)^(2n+1)`
What is `""^(47)C_(4) + ""^(51)C_(3)+sum_(n=2)^(5)""^(52-n)C_(3)` equal to ?

Options

A. `""^(52)C_(4)`
B. `""^(51)C_(5)`
C. `""^(53)C_(4)`
D. `""^(52)C_(5)`

Answer

To solve the problem, we need to evaluate the expression \( \binom{47}{4} + \binom{51}{3} + \sum_{n=2}^{5} \binom{52-n}{3} \).

### Step-by-Step Solution:

1. **Evaluate \( \binom{47}{4} \)**:
- This is a straightforward binomial coefficient calculation. It represents the number of ways to choose 4 items from 47.
- \( \binom{47}{4} = \frac{47!}{4!(47-4)!} \)

2. **Evaluate \( \binom{51}{3} \)**:
- Similarly, calculate the binomial coefficient for choosing 3 items from 51.
- \( \binom{51}{3} = \frac{51!}{3!(51-3)!} \)

3. **Evaluate the summation \( \sum_{n=2}^{5} \binom{52-n}{3} \)**:
- We will compute the individual terms of the summation:
- For \( n=2 \): \( \binom{52-2}{3} = \binom{50}{3} \)
- For \( n=3 \): \( \binom{52-3}{3} = \binom{49}{3} \)
- For \( n=4 \): \( \binom{52-4}{3} = \binom{48}{3} \)
- For \( n=5 \): \( \binom{52-5}{3} = \binom{47}{3} \)

4. **Combine the results**:
- Now we can write the summation explicitly:
\[ \sum_{n=2}^{5} \binom{52-n}{3} = \binom{50}{3} + \binom{49}{3} + \binom{48}{3} + \binom{47}{3} \]

5. **Use the Hockey Stick Identity**:
- The Hockey Stick Identity states that:
\[ \binom{r}{k} + \binom{r+1}{k} + \binom{r+2}{k} + \ldots + \binom{n}{k} = \binom{n+1}{k+1} \]
- Applying this to our summation:
\[ \binom{50}{3} + \binom{49}{3} + \binom{48}{3} + \binom{47}{3} = \binom{51}{4} \]

6. **Combine all parts**:
- Now we can combine everything we have calculated:
\[ \binom{47}{4} + \binom{51}{3} + \binom{51}{4} \]

7. **Use the Hockey Stick Identity again**:
- Combine \( \binom{51}{3} + \binom{51}{4} \):
\[ \binom{51}{3} + \binom{51}{4} = \binom{52}{4} \]

8. **Final result**:
- Thus, we have:
\[ \binom{47}{4} + \binom{52}{4} \]

9. **Final answer**:
- The final answer is \( \binom{52}{4} \).