The value of `""^50 C_4+sum_(r=1)^6 "^(56-r) C_3=`

Options

A. `"^56 C_3`
B. `"^56C_4`
C. `"^55C_4`
D. `"^55C_3`

Answer

To solve the expression \( \binom{50}{4} + \sum_{r=1}^{6} \binom{56-r}{3} \), we will break it down step by step.

### Step 1: Understanding the Summation
The summation \( \sum_{r=1}^{6} \binom{56-r}{3} \) can be expanded for each value of \( r \) from 1 to 6:
- For \( r=1 \): \( \binom{55}{3} \)
- For \( r=2 \): \( \binom{54}{3} \)
- For \( r=3 \): \( \binom{53}{3} \)
- For \( r=4 \): \( \binom{52}{3} \)
- For \( r=5 \): \( \binom{51}{3} \)
- For \( r=6 \): \( \binom{50}{3} \)

Thus, we can rewrite the summation as:
\[ \sum_{r=1}^{6} \binom{56-r}{3} = \binom{55}{3} + \binom{54}{3} + \binom{53}{3} + \binom{52}{3} + \binom{51}{3} + \binom{50}{3} \]

### Step 2: Combining the Terms
Now, we can combine this with \( \binom{50}{4} \):
\[ S = \binom{50}{4} + \left( \binom{55}{3} + \binom{54}{3} + \binom{53}{3} + \binom{52}{3} + \binom{51}{3} + \binom{50}{3} \right) \]

### Step 3: Applying the Hockey Stick Identity
We can use the hockey stick identity in combinatorics, which states:
\[ \sum_{i=k}^{n} \binom{i}{k} = \binom{n+1}{k+1} \]
In our case, we can apply this identity to the terms \( \binom{55}{3} + \binom{54}{3} + \binom{53}{3} + \binom{52}{3} + \binom{51}{3} + \binom{50}{3} \):
\[ \sum_{i=50}^{55} \binom{i}{3} = \binom{56}{4} \]

### Step 4: Final Expression
Now substituting this back into our expression for \( S \):
\[ S = \binom{50}{4} + \binom{56}{4} \]

### Step 5: Using the Pascal's Rule
Using Pascal's rule, we know:
\[ \binom{n}{r} + \binom{n+1}{r} = \binom{n+1}{r+1} \]
Thus, we can rewrite:
\[ S = \binom{56}{4} + \binom{56}{4} = 2 \cdot \binom{56}{4} \]

### Final Answer
Therefore, the final value of the expression is:
\[ S = 2 \cdot \binom{56}{4} \]