The value of expression`""^(47) C_4 +sum_(j=1)^(5) "^(52-j) C_3` is equal to

Options

A. `"^47C_5`
B. `"^52C_5`
C. `"^52C_4`
D. none of these

Answer

To solve the expression \( \binom{47}{4} + \sum_{j=1}^{5} \binom{52-j}{3} \), we will break it down step by step.

### Step 1: Rewrite the Summation
The summation \( \sum_{j=1}^{5} \binom{52-j}{3} \) can be expanded as follows:
- For \( j=1 \): \( \binom{51}{3} \)
- For \( j=2 \): \( \binom{50}{3} \)
- For \( j=3 \): \( \binom{49}{3} \)
- For \( j=4 \): \( \binom{48}{3} \)
- For \( j=5 \): \( \binom{47}{3} \)

Thus, we can write:
\[ \sum_{j=1}^{5} \binom{52-j}{3} = \binom{51}{3} + \binom{50}{3} + \binom{49}{3} + \binom{48}{3} + \binom{47}{3} \]

### Step 2: Combine the Terms
Now, we can rewrite the entire expression:
\[ \binom{47}{4} + \binom{51}{3} + \binom{50}{3} + \binom{49}{3} + \binom{48}{3} + \binom{47}{3} \]

### Step 3: Apply the Combination Identity
We can use the identity \( \binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1} \) to simplify the terms.

1. Start with \( \binom{51}{3} + \binom{50}{3} = \binom{52}{4} \)
2. Then, \( \binom{49}{3} + \binom{48}{3} = \binom{50}{4} \)
3. Finally, \( \binom{47}{3} \) remains as is.

Now we can rewrite:
\[ \binom{47}{4} + \binom{52}{4} + \binom{50}{4} + \binom{47}{3} \]

### Step 4: Combine Further
Next, we can combine \( \binom{47}{4} + \binom{47}{3} \):
\[ \binom{47}{4} + \binom{47}{3} = \binom{48}{4} \]

So now we have:
\[ \binom{48}{4} + \binom{52}{4} + \binom{50}{4} \]

### Step 5: Apply the Combination Identity Again
Now we can combine \( \binom{52}{4} + \binom{50}{4} \):
\[ \binom{52}{4} + \binom{50}{4} = \binom{51}{5} \]

### Final Step: Combine All Terms
Now we have:
\[ \binom{48}{4} + \binom{51}{5} \]

This cannot be simplified further using the combination identities, so we can leave it in this form.

### Final Answer
The value of the expression is:
\[ \binom{48}{4} + \binom{51}{5} \]