The value of ` ""^(50)C_4 + underset(r=1) overset(6)sum ""^(56-r)C_3 ` is

Options

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

Answer

To solve the problem \( {}^{50}C_4 + \sum_{r=1}^{6} {}^{56-r}C_3 \), we will follow these steps:

### Step 1: Understand the Summation
The summation \( \sum_{r=1}^{6} {}^{56-r}C_3 \) can be expanded as follows:
- When \( r = 1 \), it gives \( {}^{55}C_3 \)
- When \( r = 2 \), it gives \( {}^{54}C_3 \)
- When \( r = 3 \), it gives \( {}^{53}C_3 \)
- When \( r = 4 \), it gives \( {}^{52}C_3 \)
- When \( r = 5 \), it gives \( {}^{51}C_3 \)
- When \( r = 6 \), it gives \( {}^{50}C_3 \)

Thus, the summation can be rewritten as:
\[ {}^{55}C_3 + {}^{54}C_3 + {}^{53}C_3 + {}^{52}C_3 + {}^{51}C_3 + {}^{50}C_3 \]

### Step 2: Combine the Terms
Now, we can combine the original term with the summation:
\[ {}^{50}C_4 + {}^{55}C_3 + {}^{54}C_3 + {}^{53}C_3 + {}^{52}C_3 + {}^{51}C_3 + {}^{50}C_3 \]

### Step 3: Apply the Hockey Stick Identity
We can use the Hockey Stick Identity in combinatorics, which states:
\[ {}^{n}C_{r} + {}^{n+1}C_{r} + {}^{n+2}C_{r} + \ldots + {}^{m}C_{r} = {}^{m+1}C_{r+1} \]
In our case, we can apply it to the terms from \( {}^{50}C_3 \) to \( {}^{55}C_3 \):
\[ {}^{51}C_3 + {}^{52}C_3 + {}^{53}C_3 + {}^{54}C_3 + {}^{55}C_3 = {}^{56}C_4 \]

### Step 4: Substitute Back
Now, we can substitute this back into our equation:
\[ {}^{50}C_4 + {}^{56}C_4 \]

### Step 5: Combine the Terms
Now we can combine the two terms:
\[ {}^{50}C_4 + {}^{56}C_4 = {}^{57}C_4 \]

### Final Answer
Thus, the value of the expression \( {}^{50}C_4 + \sum_{r=1}^{6} {}^{56-r}C_3 \) is:
\[ {}^{57}C_4 \]

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