MHT CET · Maths · Binomial Theorem
The value of \({ }^{47} C_4+\sum_{j=1}^5{ }^{(52-j)} C_3\) is
- A \({ }^{52} \mathrm{C}_4\)
- B \({ }^{52} \mathrm{C}_2\)
- C \({ }^{48} \mathrm{C}_4\)
- D \({ }^{48} \mathrm{C}_2\)
Answer & Solution
Correct Answer
(A) \({ }^{52} \mathrm{C}_4\)
Step-by-step Solution
Detailed explanation
\( { }^{47} C_4+\sum_{j=1}^5{ }^{(52-j)} C_3 = { }^{47} C_4 + { }^{47} C_3 + { }{ }^{48} C_3 + { }{ }^{49} C_3 + { }{ }^{50} C_3 + { }{ }^{51} C_3 \) \( = { }{ }^{48} C_4 + { }{ }^{48} C_3 + { }{ }^{49} C_3 + { }{ }{ }^{50} C_3 + { }{ }^{51} C_3 \)
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