If `""^(47)C_(4)+sum_(r=1)^(5) ""^(52-r)C_(3)` is equal to

Options

A. `.^(47)C_(6)`
B. `.^(52)C_(5)`
C. `.^(52)C_(4)`
D. None of these

Answer

To solve the problem \( \binom{47}{4} + \sum_{r=1}^{5} \binom{52-r}{3} \), we will use the properties of combinations, specifically the identity that relates combinations of consecutive integers.

### Step-by-step Solution:

1. **Understanding the Sum**:
We start with the expression \( \sum_{r=1}^{5} \binom{52-r}{3} \). This means we will calculate the values of \( \binom{52-1}{3}, \binom{52-2}{3}, \binom{52-3}{3}, \binom{52-4}{3}, \) and \( \binom{52-5}{3} \).

2. **Calculating Each Term**:
- For \( r = 1 \): \( \binom{51}{3} \)
- For \( r = 2 \): \( \binom{50}{3} \)
- For \( r = 3 \): \( \binom{49}{3} \)
- For \( r = 4 \): \( \binom{48}{3} \)
- For \( r = 5 \): \( \binom{47}{3} \)

Thus, we can rewrite the sum as:
\[ \sum_{r=1}^{5} \binom{52-r}{3} = \binom{51}{3} + \binom{50}{3} + \binom{49}{3} + \binom{48}{3} + \binom{47}{3} \]

3. **Applying the Combination Identity**:
We will use the identity:
\[ \binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1} \]
to combine the terms.

4. **Combining Terms**:
- Start with \( \binom{51}{3} + \binom{50}{3} \):
\[ \binom{51}{3} + \binom{50}{3} = \binom{52}{4} \]
- Now, add \( \binom{49}{3} \):
\[ \binom{52}{4} + \binom{49}{3} = \binom{50}{4} \]
- Next, add \( \binom{48}{3} \):
\[ \binom{50}{4} + \binom{48}{3} = \binom{49}{4} \]
- Finally, add \( \binom{47}{3} \):
\[ \binom{49}{4} + \binom{47}{3} = \binom{48}{4} \]

5. **Final Calculation**:
Now we have:
\[ \sum_{r=1}^{5} \binom{52-r}{3} = \binom{48}{4} \]
Therefore, the original expression becomes:
\[ \binom{47}{4} + \binom{48}{4} \]
Applying the combination identity again:
\[ \binom{47}{4} + \binom{48}{4} = \binom{49}{5} \]

6. **Conclusion**:
Thus, the final result is:
\[ \binom{49}{5} \]