Question

{"given":"$$^{47}C_4 + \\sum_{j=1}^{5} {}^{(52-j)}C_3 = {}^{47}C_4 + {}^{51}C_3 + {}^{50}C_3 + {}^{49}C_3 + {}^{48}C_3 + {}^{47}C_3$$","key_observation":"Apply Pascal's identity repeatedly: $^nC_r + {}^nC_{r-1} = {}^{n+1}C_r$. Group $^{47}C_4 + {}^{47}C_3$ first, then continue combining.","option_analysis":[{"label":"(A)","text":"$^{47}C_5$","verdict":"incorrect","explanation":"The telescoping application of Pascal's rule leads to $^{52}C_4$, not $^{47}C_5$. The index increases with each application, so $^{47}C_5$ is not achievable here."},{"label":"(B)","text":"$^{52}C_5$","verdict":"incorrect","explanation":"The lower index remains 4 (not 5) throughout the telescoping process, since we started with $^{47}C_4$ and applied $^nC_r + {}^nC_{r-1} = {}^{n+1}C_r$ five times."},{"label":"(C)","text":"$^{52}C_4$","verdict":"correct","explanation":"Step-by-step: $({}^{47}C_4 + {}^{47}C_3) = {}^{48}C_4$; $({}^{48}C_4 + {}^{48}C_3) = {}^{49}C_4$; $({}^{49}C_4 + {}^{49}C_3) = {}^{50}C_4$; $({}^{50}C_4 + {}^{50}C_3) = {}^{51}C_4$; $({}^{51}C_4 + {}^{51}C_3) = {}^{52}C_4$."},{"label":"(D)","text":"None of these","verdict":"incorrect","explanation":"The answer $^{52}C_4$ is option (C), so 'None of these' is incorrect."}],"answer":"(C)","formula_steps":[]}