Question

The value of the expression \[ {}^{47}C_4 + \sum_{j=1}^{5} {}^{52-j}C_3 \] is:

• A. \( {}^{52}C_3 \)
• B. \( {}^{51}C_4 \)
• C. \( {}^{52}C_4 \)
• D. \( {}^{51}C_3 \)

Hint:

For binomial sums: \begin{itemize} \item Use Pascal triangle identities. \item Convert sums into telescoping forms. \end{itemize}

Answer 1

The Correct Option is C

Concept: Use identity: \[ {}^nC_r + {}^{n-1}C_r + \cdots = {}^{n+1}C_{r+1} \] Also Pascal identity: \[ {}^nC_r + {}^nC_{r+1} = {}^{n+1}C_{r+1} \] Step 1: {\color{red}Expand summation.} \[ \sum_{j=1}^{5} {}^{52-j}C_3 = {}^{51}C_3 + {}^{50}C_3 + {}^{49}C_3 + {}^{48}C_3 + {}^{47}C_3 \] Step 2: {\color{red}Combine with given term.} Expression becomes: \[ {}^{47}C_4 + ({}^{47}C_3 + {}^{48}C_3 + \cdots + {}^{51}C_3) \] Use identity: \[ {}^nC_4 = {}^{n-1}C_3 + {}^{n-1}C_4 \] So combining telescoping terms leads to: \[ {}^{52}C_4 \] Step 3: {\color{red}Final result.} \[ \boxed{{}^{52}C_4} \]