Question

The value of \[ {}^{47}C_{4} + \sum_{j=1}^{5} {}^{(52-j)}C_{3} \] is:

• A. \( {}^{52}C_4 \)
• B. \( {}^{52}C_2 \)
• C. \( {}^{48}C_4 \)
• D. \( {}^{48}C_2 \)

Hint:

Always start combining terms with the smallest index using the Pascal's identity: ${}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks for the sum of a combination term and a summation series. We use the identity \( {}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r \).
Step 3: Detailed Explanation:
The given expression is:
\[ S = {}^{47}C_4 + \sum_{j=1}^5 {}^{(52-j)}C_3 \]
Expanding the summation:
\[ S = {}^{47}C_4 + ({}^{51}C_3 + {}^{50}C_3 + {}^{49}C_3 + {}^{48}C_3 + {}^{47}C_3) \]
Rearranging to pair terms from smallest \( n \):
\[ S = ({}^{47}C_4 + {}^{47}C_3) + {}^{48}C_3 + {}^{49}C_3 + {}^{50}C_3 + {}^{51}C_3 \]
Applying the identity \( {}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r \):
Step 1: \( {}^{47}C_4 + {}^{47}C_3 = {}^{48}C_4 \)
Step 2: \( {}^{48}C_4 + {}^{48}C_3 = {}^{49}C_4 \)
Step 3: \( {}^{49}C_4 + {}^{49}C_3 = {}^{50}C_4 \)
Step 4: \( {}^{50}C_4 + {}^{50}C_3 = {}^{51}C_4 \)
Step 5: \( {}^{51}C_4 + {}^{51}C_3 = {}^{52}C_4 \)
Step 4: Final Answer:
The final sum is \( {}^{52}C_4 \).