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: Expand the Sum
Sum $= {}^{51}C_3 + {}^{50}C_3 + {}^{49}C_3 + {}^{48}C_3 + {}^{47}C_3$.

Step 2: Combine with Initial Term
Start with ${}^{47}C_4 + {}^{47}C_3$.
Using ${}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r$, we get ${}^{48}C_4$.

Step 3: Recursive Application
${}^{48}C_4 + {}^{48}C_3 = {}^{49}C_4$
${}^{49}C_4 + {}^{49}C_3 = {}^{50}C_4$
${}^{50}C_4 + {}^{50}C_3 = {}^{51}C_4$
${}^{51}C_4 + {}^{51}C_3 = {}^{52}C_4$.

Step 4: Conclusion
The final result is ${}^{52}C_4$.
Final Answer: (A)